10th – Science – PTA Question – 1: Answer Key

PTA Question - 1 Answer key

PTA – Model question Paper – 1 PTA – khjphp tpdhj;jhs; – 1
PART – I gFjp – I
1. c) change of momentum 1. ,) ce;j khw;wk;
2. c) remains same 2. ,) khw;wkpy;iy
3. d) glass > water > air 3. <) fz;zhb > ePh; > fhw;W
4. c) carbon dioxide 4. ,) fhh;gd;ilMf;irL
5. a) 17th 5. m) 17tJ
6. b) deliquescence 6. m) <uk; cwpQ;rpf;fiujy;
7. d) canine 7. <) Nfhiug; gy;
8. b) thalamus 8. M) jyhk];
9. a) mitosis 9. m) ikl;lhrp];
10. a) radio-carbon method 10. m) NubNah fhh;gd; Kiw
11. a) May 31 11. m) Nk 31
12. a) script area 12. m) Script area
PART –II gFjp – II
13. Unit-2 : TB Pg.no: 30 WTS Guide Pg.no: 25 13. myF-2: Gj;jf vz;: 30 WTS g.vz; : 26
14. Unit-6 : TB Pg.no: 89 WTS Guide Pg.no: 85 14. myF-6: Gj;jf vz;: 88 WTS g.vz; : 88
15. Unit-10 : TB Pg.no: 153 WTS Guide Pg.no: 130 15. myF-10: Gj;jf vz;: 151 WTS g.vz; : 134
16. Unit-11 : TB Pg.no: 171 WTS Guide Pg.no: 141 16. myF-11: Gj;jf vz;: 167 WTS g.vz; : 147
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17. Unit-12 : (Inside TB.Pg.no: 174) Vascular bundle : 17. myF-12: Inside TB.Pg.no: 171
Xylem and phloem tissues are present in the form of th];Fyhu; fw;iw : iryk; kw;Wk; GNshak;
jpRf;fisf; nfhz;Ls;s fw;iwfs; th];FyH
bundles called vascular bundles. Xylem conducts fw;iwfs; vdg;gLk;. iryk; ePu; kw;Wk;
water and minerals. Phloem conducts food fdpkq;fis flj;JfpwJ. GNshak; czTg;
materials. nghUs;fis flj;JfpwJ.
18. myF-14: Gj;jf vz;: 211 WTS g.vz; : 183
18. Unit-14 : TB Pg.no: 215 WTS Guide Pg.no: 177
19. myF-16: cs;Gj;jf vz;: 232
19. Unit-16 : Inside Pg.no: 235

The person shown in the figure is suffering from glj;jpy; cs;s egiug; ghjpj;Js;s FiwghL
goitre. It is caused due too the inadequate supply of vspa fha;l;lu;. cztpy; Njitahd msT
iodine in our diet. mNahbd; ,y;yhjjhy; Vw;gLfpwJ.
20. Unit-18 : TB Pg.no: 272 WTS Guide Pg.no: 225 20. myF-18: Gj;jf vz;: 271 WTS g.vz; : 237

21. Unit-22 : TB Pg.no: 327 WTS Guide Pg.no: 266 21. myF-22: Gj;jf vz;: 327 WTS g.vz; : 281
22. Unit-1 : Given data : 22. myF-1: nfhLf;fg;gl;lit:
-2
Acceleration (a) = 1.8 ms m= 50 kg KLf;fk; = 1.8 kPtp-2 epiw = 50 fpfp> kpd;J}f;fp
If Lift is moving downward with an acceleration ‘a’ vd;w KLf;f kjpg;gpy; fPNo efu;fpwJ vdpy;>
'a' then the Apparent weight is , R= m(g-a) Njhw;w vil
R = 50(9.8 – 1.8) = 50 × 8 R= m(g-a)= R = 50(9.8 – 1.8) = 50 × 8
So, Apparent weight is = 400 N Njhw;w vil = 400 N

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10th – Science – PTA Question – 1: Answer Key
PART –III gFjp – III
23. Unit-1 : TB Pg.no: 14 WTS Guide Pg.no: 8 23. myF-1: Gj;jf vz;: 13 WTS g.vz; : 8
24. Unit-3 : i) The ratio of increase in length of the body 24. myF-3: i) XuyF ntg;gepiy cau;thy; nghUspd;
per degree rise in temperature to its unit length is ePsj;jpy; Vw;gLk; khw;wj;jpw;Fk; XuyF ePsj;jpw;Fk;
called as the coefficient of linear expansion. cs;s jfT ePs; ntg;g tpupT Fzfk; MFk;.
Unit-3 : ii) ΔL= increase in length, ∆T = 323K- 303K = 20 K myF-3: ii) ΔL= ePsj;jpy; Vw;gLk; khw;wk;>
L0 (Original length)= 50m, αL =23 ×10-6K-1 ∆T = 323K- 303K = 20 K, αL =23 ×10-6K-1
∆𝐿 𝐿0 (cz;ikahd ePsk;)= 50m
=αL ∆T ∆𝐿
𝐿0
=αL ∆T
𝐿0
∆𝐿 = αL ∆T × 𝐿0 = (23 ×10-6)×20 × 50 = 0.023
∆𝐿 = αL ∆T × 𝐿0 = (23 ×10-6) × 20 × 50= 0.023
25. Unit-5 : i) and ii) TB Pg.no: 72 WTS Pg.no: 75 25. myF-5: i) kw;Wk; ii) Gj;jf vz;:72 WTSg.vz;:77

26. Unit-7 : i) TB Pg.no: 104 WTS Guide Pg.no: 100 26. myF-7: i)Gj;jf vz;: 103 WTS g.vz; : 102
Unit-8 : ii) TB Pg.no: 122 WTS Guide Pg.no: 111 myF-8: ii)Gj;jf vz;: 119 WTS g.vz; : 114

27. Unit-9 : i) TB Pg.no: 135 WTS Guide Pg.no: 120 27.myF-9: i)Gj;jf vz;: 132 WTS g.vz; : 124
Unit-10 :ii) TB Pg.no: 154 WTS Guide Pg.no: 133 myF-10: ii)Gj;jf vz;: 151 WTS g.vz; : 138

28. Unit-20: i) TB Pg.no: 298 WTS Guide Pg.no: 246 28.myF-20: i)Gj;jf vz;: 297 WTS g.vz; : 260
Unit-21 :ii) TB Pg.no: 312 WTS Guide Pg.no: 255 myF-21: ii)Gj;jf vz;: 313 WTS g.vz; : 270

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29. Unit-15 : TB Pg.no: 228 WTS Guide Pg.no: 191 29.myF-15: Gj;jf vz;: 225 WTS g.vz; : 197

30. Unit-13: i) TB Pg.no: 198 WTS Guide Pg.no: 164 30.myF-13: i)Gj;jf vz;: 195 WTS g.vz; : 171
Unit-17 :ii) TB Pg.no: 258 WTS Guide Pg.no: 215 myF-17: ii)Gj;jf vz;: 255 WTS g.vz; : 225

31. Unit-19 : TB Pg.no: 283 WTS Guide Pg.no: 236 31.myF-19: Gj;jf vz;: 283 WTS g.vz; : 250

32. Unit-8: i) 18th group elements (or) ‘0’ group 32.myF-8: i) 18tJ njhFjpj; jdpkq;fs; (my;yJ)
G+[;[paj; njhFjpj; jdpkq;fs;- He, Ne, Ar, Kr
elements - He, Ne, Ar, Kr
myF-11: ii)
Unit-11: ii) <j;jPdpy; typikFiwe;j ,ul;il Ntjpg;gpizg;G
Ethene has a double bond, which is ,Ug;gjhy; mjd; epiyg;Gj;jd;ik <j;Njidtplf;
comparatively unstable when compared to FiwT. vdNt <j;Njidtpl <j;jPd; tpidjpwd;
ethane. Thus ethene is more reative than ethane. kpf;fJ.
gFjp – IV
PART –IV
33A. Unit-2: i) 33A. myF-2: i)

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10th – Science – PTA Question – 1: Answer Key
33A. ii) Unit-2: Name of the defect - Myopia also 33A. ii) myF-2: Fiwghl;bd; ngau; - ikNahgpah
known as short sightedness occurs due to the vd;gJ fpl;lg;ghu;it vdg;gLk;. ,J tpopf;Nfhsk;
rpwpJ ePz;L tpLtjhy; Vw;gLfpwJ. ,jdhy; mUfpy;
lengthening of eyeball. With this defect nearby
cs;s nghUs;fisj; njspthf fhzKbAk;. Mdhy;
objects can be seen clearly but distant objects cannot njhiytpy; cs;s nghUs;fis fhzKbahJ.
be seen clearly. This defect can be corrected using a jFe;j Ftpaj; njhiyT nfhz;l Fopnyd;irg;
concave lens. gad;gLj;jp rupnra;ayhk;.

Unit-6: iii) myF-6: iii)

Number of neutrons epA+l;uhd;fspd; vz;zpf;if

= mass number - atomic number = epiw vz; - mZ vz;
= 231 – 90 = 141 = 231 – 90 = 141

33B.Unit-4:i)ii)iii) TB Pg.no: 57 WTS Guide Pg.no:56 33B.myF-4:i),ii),iii) Gj;jf vz;: 58 WTSg.vz; : 57

Unit-4: iv) TB Pg.no: 57 WTS Guide Pg.no:57 myF-4: iv) Gj;jf vz;: 58 WTS g.vz; : 58

34A.Unit-8: i) TB Pg.no: 123 WTS Guide Pg.no: 112 34A. myF-8: i)Gj;jf vz;: 119 WTS g.vz; : 115
Unit-11:ii)TB Pg.no: 171 WTS Guide Pg.no: 144 myF-11: ii) Gj;jf vz;: 168 WTS g.vz; : 151

34B.Unit-9: i) “Like solvent dissolves like solute” 34B. myF-9: i) “xj;j fiunghUl;fs; xj;j
This expression means that dissolving occurs fiug;ghdpy; fiufpwJ”
fiunghUSf;Fk; fiug;ghDf;Fk; ,ilNa xw;Wik
when similarities exist between the solvent and fhzg;gLk; NghJ jhd; fiujy; epfo;fpwJ.
the solute. KidTWk;; Nru;kq;fs; KidTWk; fiug;ghdpy;
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Polar Compounds are soluble in polar solvents-
Ex: Common salt is a polar compound and
vspjpy; fiufpwJ.
v.fh : rikay; cg;G KidTWk; Nru;kk; vdNt
,J KidTWk; fiug;ghdhd ePupy; vspjpy;
dissolves readily in polar solvent like water. fiufpwJ.
Non-polar compounds are soluble in non-polar KidTwhr; Nru;kq;fs; KidTwh fiug;ghdpy;
solvents - Ex: Fat dissolved in ether. But non- vspjpy; fiufpwJ.
v.fh : <jupy; fiuf;fg;gl;l nfhOg;G. Mdhy;>
polar compounds, do not dissolve in polar KidTwhr; Nru;kq;fs; KidTWk; fiug;ghdpy;
solvents; polar compounds do not dissolve in fiutjpy;iy. mJNghy> KidTWk; Nru;kq;fs;
non-polar solvents. KidTwh fiug;ghdpy; fiutjpy;iy.
34B.ii) Unit-10 : 34.B. ii) myF-10 :
Decomposition :2 KClO3 ⟶ 2KCl + 3O2 rpijT tpid : 2 KClO3 ⟶ 2KCl + 3O2
Potassium chlorate is decomposed as nghl;lhrpak; FNshNul; Nru;kkhdJ
Potassium chloride and Oxygen. nghl;lhrpak; FNshiuL kw;Wk; Mf;]p[d;
%yf;$Wfshf rpijtilfpwJ.
Single Displacement:Zn+CuSO4⟶ ZnSO4+ Cu
xw;iw ,lg;ngau;r;rp: Zn+CuSO4⟶ ZnSO4+ Cu
Zinc element displaces copper in copper [pq;f; jdpkk; fhg;gu; ry;/Ngl;bypUe;J fhg;giu
sulphate and forms zinc sulphate and elemental ,lg;ngau;r;rp nra;J [pq;f; ry;Ngl; kw;Wk;
copper. fhg;gu; jdpkj;ij cUthf;FfpwJ.
Combination: 2 Mg + O2 ⟶ 2MgO Nru;f;if tpid: 2 Mg + O2 ⟶ 2MgO
Magnesium combines with Oxygen to form nkf;dPrpak;> Mf;]p[Dld; ,ize;J
Magnesium Oxide. nkf;dPrpak; Mf;i]ilj; jUfpwJ.
Double Displacement: ,ul;il ,lg;ngau;r;rp:
Na2SO4 + BaCl2 ⟶ BaSO4 + 2NaCl Na2SO4 + BaCl2 ⟶ BaSO4 + 2NaCl
Sodium and barium interchange its position to Nrhbak; kw;Wk; Ngupak; madpfs;
,lngau;r;rpahtjhy; Ngupak; ry;Ngl; kw;Wk;
form barium sulphate and sodium chloride. Nrhbak; FNshiuL cUthfpwJ.

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10th – Science – PTA Question – 1: Answer Key
35A.Unit-12: i) TB Pg.no:185 WTS Guide Pg.no:158 35A. myF-12: i) WTS g.vz; : 164
Unit-14:ii)TB Pg.no: 214 WTS Guide Pg.no: 176 myF-14: ii) Gj;jf vz;: 211 WTS g.vz; : 182
Unit-20:iii)TB Pg.no: 298 WTS Guide Pg.no:247 myF-20: iii) Gj;jf vz;: 297 WTS g.vz; :260

35B.Unit-22:i) The 3R approach such as Reduce, 35B. myF-22: i) fopTfis rpwg;ghd Kiwapy;
Reuse and Recycle may be followed for ifahStjw;F 3R Kiw Vw;wjhFk;. Reduce
effective waste management. -Fiwj;jy;> Reuse - kWgad;ghL> Recycle -
kWRow;rp.
Unit-18:ii) myF-18: ii)
a) Helicase - Inside TB Pg.no:268 m) n`ypNf]; - Gj;jf vz;: 266
b) Topoisomerase - Inside TB Pg.no:268 M) NlhNgh[NrhnkNu]; - Gj;jf vz;: 266
c)DNA Polymerase- Inside TB Pg.no:268 ,) b.vd;.V ghypnkNu]; - Gj;jf vz;: 267
d) DNA ligases - Inside TB Pg.no:268 <) b.vd;.V iyNf]; - Gj;jf vz;: 267
e) TerminusInside - Inside TB Pg.no:268 c) nlh;kpd]; - Gj;jf vz;: 267

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10th – Science – PTA Question – 2: Answer Key
PTA Question - 2 Answer key
PTA – Model question Paper – 2 PTA – khjphp tpdhj;jhs; – 2
PART – I gFjp – I
1. d) bifocal lens 1. <) ,UFtpa nyd;R
2. b) 10 Ω (Hint : R=5+2+3 =10 Ω) 2. M) 10 Ω (Hint : R=5+2+3 =10 Ω)
3. ,) GNuhl;lhd;fspd; vz;zpf;ifapy; xd;W
3. c) number of proton increases by one
mjpfupf;Fk; (tpdhj;jhspy; jtwhf FiwAk;
4. a) thermal expansion of liquids vd nfhLf;fg;gl;Ls;sJ)
5. b) A- III, B-IV, C-II, D-I 4. m) jputq;fspd; ntg;gtphpT
6. c) oxygen gas is utilised 5. M) A- III, B-IV, C-II, D-I
6. ,) Mf;rp[d; thA gad;gLj;jg;gLfpwJ
7. b) A is xylem, B is phloem
7. M) A - iryk;> B - GNshak;
8. b) pituitary gland 8. M) gpl;A+l;lhp Rug;gp
9. a) secondary constriction 9. m) ,uz;lhk;epiyr; RUf;fk;
10. b) restriction endonucleases 10. M) nu];l;hpf;]d; vz;NlhepA+f;spNa];
11. a) hydropower 11. m) ePh; Mw;wy;
12. M) Script editor
12. b) Script editor
PART –II gFjp – II
13. Unit-3 :TB Pg.no:40 WTS Guide Pg.no: 40 13. myF-3: Gj;jf vz;: 41 WTS g.vz; : 41

14. Unit-4 : Two 5Ω resistors are connected in series 14. myF-4: ,uz;L 5Ω kpd;jilfs; njhlu;
Ω
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which is parallel to 10 Ω resistor.
Rs = 5 + 5 = 10
,izg;gpy; ,izf;fg;gl;L gpd;> 10
kpd;jilAld; ,izahf ,izf;fg;gl;Ls;sJ.
1 1 2 1 Rs = 5 + 5 = 10
Rp = + = = = 5Ω 1 1 2 1
10 10 10 5
Rp = + = = = 5Ω
10 10 10 5
15. Unit-5 : TB Pg.no: 71 WTS Guide Pg.no: 71 15. myF-5: fhw;W %yf;$Wfspd; mjph;Tfspd; tPr;R
jput %yf;$Wfis tpl mjpfkhf cs;sJ.
mjdhy; epuk;gpa ghj;jpuj;ijtpl fhypahf cs;s
ghj;jpuj;jpy; xypahdJ mjpf msT cUthfpwJ.
16. Unit-8 : TB Pg.no: 122 WTS Guide Pg.no: 111 16. myF-8: Gj;jf vz;: 119 WTS g.vz; : 114

17. Unit-13 :TB.Pg.no: 198 WTS Guide Pg.no: 165 17. myF-13: Gj;jf vz;: 195 WTS g.vz; : 172

18. Unit-15 : TB Pg.no: 227 WTS Guide Pg.no: 189 18. myF-15: Gj;jf vz;: 224 WTS g.vz; : 195

19. Unit-17: TB Pg.no: 258 WTS Guide Pg.no:216 19. myF-17: Gj;jf vz;: 255 WTS g.vz; : 225

20. Unit-19 : TB Pg.no: 283 WTS Guide Pg.no: 234 20. myF-19: Gj;jf vz;: 283 WTS g.vz; : 248
21. Unit-21 : TB Pg.no: 312 WTS Guide Pg.no: 255 21. myF-21: Gj;jf vz;: 313 WTS g.vz; : 269
22.Unit-9 : 22. myF-9:
Compound A - Magnesium sulphate heptahydrate Nrh;kk; A - nkf;dr P pak; ry;Ngl; n`g;lhi`l;Nul;
On heating it becomes an anhydrous salt MgSO4 ntg; g g; g Lj; Jk; NghJ 7 ePu; %yf;$Wfis ,oe;J
by losing 7 water molecules. MgSO 4 vd; w eP uw; w cg;ghf khWfpwJ. gr;ir
Green vitriol - FeSO4. 7H2O having 7 water tpl;upahpypy; cs;s – (FeSO4. 7H2O) 7 ePu;
molecules which is equal to number of water %yf;$Wfspd; vz;zpf;ifAk; Nru;kk; A ,oe;j
molecules lost by A. ePu; %yf;$Wfspd; vz;zpf;ifAk; rkk;.

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10th – Science – PTA Question – 2: Answer Key
PART –III gFjp – III
23. Unit-3 : i) False. According to charles’ law at 23. myF-3: i) jtW. rhu;y]; tpjpg;gb khwh
constant pressure, temperature of the gas is directly mOj;jj;jpy; cs;s thAtpy; ntg;gepiy
proportional to its volume. gUkDf;F Neu; jftpy; mikAk;.
ii) m) ];Nfyhu; M) ntg;gepiy
ii) a) Scalar b) temperature
24. myF-4: i) Gj;jf cs;.g.vz; : 44
24. Unit-4 : i) Inside TB Pg.no: 43- a-iii, b-i, c-iv, d-ii
m-iii> M-i, ,-iv, <-ii
25. myF-7: i) Gj;jf vz;: 103 WTS g.vz; : 105
25. Unit-7 : i) TB Pg.no: 104 WTS Guide Pg.no: 103
myF-11: ii) Gj;jf vz;: 167 WTS g.vz; : 147
Unit-11 : ii) TB Pg.no: 171 WTS Guide Pg.no:141

26. Unit-9 : i) TB Pg.no: 135 WTS Guide Pg.no: 123 26. myF-9: i)Gj;jf vz;: 133 WTS g.vz; : 126
Unit-10 : ii) TB Pg.no: 153 WTS Guide Pg.no: 129 myF-10: ii)Gj; jf vz; : 150 WTS g.vz; : 133
m) eLepiyahf;fy; tpid
a) Neutralisation reaction M) ntg;gr;rpijT tpid
b) Thermal decomposition
27. Unit-8 : i) H = 2.1 ; Na = 1 ; F= 4 27.myF-8: i) H = 2.1 ; Na = 1 ; F= 4
Electronegativity of HF = 4 - 2.1=1.9 HF –d; vyf;l;uhd; fth; jd;ikapd; kjpg;G = 4 - 2.1=1.9
The difference is greater than 1.7, the bond is ionic 1.7[ tpl mjpfk; vdpy; mJ madpg;gpizg;G
Electronegativity of NaH = 2.1- 1= 1.1 NaH –d; vyf;l;uhd; fth; jd;ikapd; kjpg;G = 4 - 2.1=1.9
The difference is less than 1.7, the bond is Covalent 1.7[ tpl mjpfk; vdpy; mJ rfg;gpizg;G
Unit-11 :ii)a) myF-11: ii)m)

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CH3 CH2 CH2 CH2 COOH CH3 CH2 CH2 CH2 COOH
b)
M)

28. Unit-14: i)TB Pg.no: 215 WTS Guide Pg.no: 177 28.myF-14: i)Gj;jf vz;: 212 WTS g.vz; : 183

29. Unit-16 : i)TB Pg.no: 241 WTS Guide Pg.no: 202 29.myF-16: i)Gj;jf vz;: 238 WTS g.vz; : 210
Unit-18 : ii) TB Pg.no: 272 WTS Guide Pg.no: 225 myF-18: ii)Gj;jf vz;: 271 WTS g.vz; : 237

30. Unit-20: i) TB Pg.no: 298 WTS Guide Pg.no: 245 30.myF-20: i)Gj;jf vz;: 297 WTS g.vz; : 259
Unit-22 :ii) TB Pg.no: 327 WTS Guide Pg.no: 266 myF-22: ii)Gj;jf vz;: 327 WTS g.vz; : 281

31. Unit-14 : i) TB Pg.no: 215 WTS Guide Pg.no: 178 31.myF-14: i) Gj;jf vz;: 212 WTS g.vz; :184
ii) TB Pg.no: 215 WTS Guide Pg.no: 179 ii) Gj;jf vz;: 212 WTS g.vz; :185

32.myF-2: i),ii) Gj;jf cs;.g.vz; : 26 (glk; - 2.18)

32. Unit-2: i),ii) Inside TB Pg.no: 26 (Diagram 2.18)

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10th – Science – PTA Question – 2: Answer Key
PART –IV gFjp – IV
33A.Unit-1: i) For the comfort purpose, shock 33A. myF-1: i) nrhFRg; NgUe;Jfspy;
absorbers are used to absorb or damp the shocks mjph;TWQ;rpfs; Njitaw;w mjph;Tfis
or unwanted oscillations of the bus due to cwpQ;rpf;nfhz;L ek;ik ghJf;fhf;fpwJ.
damaged roads. ii) W = mg = 686 N
ii) W = mg = 686 N 𝑤 686
𝑚 = = = 70 kg
𝑤 686 𝑔 9.8
𝑚 = = = 70 kg
𝑔 9.8 W= mg=70 × 1.625 = 113.75 N
W= mg=70 × 1.625 = 113.75 N iii) WTS g.vz; : 20
iii) WTS Guide Pg.no: 19
33B.myF-5: i) Gj;jf cs;.g.vz; : 68 (Problem-2)
33B.Unit-5: i) Inside TB Pg.no: 67(Problem-2)
myF-6: ii) Gj;jf cs;.g.vz; : 82
Unit-6: ii) Inside TB Pg.no: 82

34A.Unit-7: i) TB Pg.no: 104 WTS Guide Pg.no: 102 34A. myF-7: i)Gj;jf vz;: 103 WTS g.vz; : 105
Unit-9: ii)TB Pg.no: 135 WTS Guide Pg.no: 122 myF-9: ii) Gj;jf vz;: 133 WTS g.vz; : 126
34B.Unit-11: i) 34B. myF-11: i)
Name of the Functional Nrh;kj;jpd; tpidr;nray;
Structural formula mikg;G tha;g;ghL
compound group present ngah; njhFjp
CH3 – CH – CH3 - OH CH3 – CH – CH3 - OH
Propanol GNuhg;gdhy;
OH
OH
CH3 – C – H O
Ethanal
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O –C-H vj;jNdy;
CH3 – C – H

O
O

–C-H
O
Butanone O
>C=O gpA+l;lNdhd;
CH3–CH2 –C – CH3 >C=O
CH3–CH2 –C – CH3
Butanoic CH3-CH2-CH2-COOH -COOH
gpA+l;ldhapf; CH3-CH2-CH2-COOH -COOH
acid
mkpyk;

Unit-8: ii) Second most abundant metal, A = Iron myF-8:ii) mYkpdpaj;jpw;F mLj;J> kpf mjpfkhf
When steam is passed over red hot iron, G+kpapy; fhzg;gLk; cNyhfk; A =,Uk;G MFk;.
magnetic oxide(B) is formed. nrQ;#nlw;wpa ,Uk;gpd; kPJ> ePuhtpia gha;r;Rk;
3 Fe + 4 H2O (steam) →Fe3O4 + 4 H2 ↑ NghJ Nkf;dl;bf; Mf;irL(B) cUthfpwJ.
B = Fe3O - magnetic oxide 3 Fe + 4 H2O (ePuhtp) →Fe3O4 + 4 H2 ↑
C=Nickel steel (Fe,C,Ni) it is used to make B = Fe3O - Nkf;dl;bf; Mf;irL
aircraft parts and propeller. C = epf;fy; ,Uk;G (Fe, C, Ni) ,JNt tpkhdj;jpd;
ghfq;fs; kw;Wk; ce;jpfs; jahhpg;gpy;
35A.Unit-13: i) TB Pg.no:198 WTS Guide Pg.no:164 gad;gLfpwJ.
Unit-16:ii) Inside TB Pg.no: 238 35A. myF-13:i) Gj;jf vz;: 195 WTS g.vz; : 171
Unit-17:iii) Inside TB Pg.no: 248 myF-16: ii) Gj;jf cs;.g.vz; : 235
myF-17: iii) Gj;jf cs;.g.vz; : 245
35B.Unit-19:i) TB Pg.no: 284 WTS GuidePg.no:237 35B. myF-19:i) Gj;jf vz;: 283 WTS g.vz; : 250
Unit-22:ii)TB Pg.no: 327 WTS GuidePg.no:269 myF-22:ii) Gj;jf vz;: 328 WTS g.vz; : 284

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10th – Science – PTA Question – 3: Answer Key
PTA Question - 3 Answer key
PTA – Model question Paper – 3 PTA – khjphp tpdhj;jhs; – 3
PART – I gFjp – I
1. c) a parallel beam of light 1. ,) ,izf;fw;iwfis cUthf;Fk;
2. d) all the above 2. <) ,it midj;Jk;
3. ,) fhhPak;
3. c) lead
4. ,) 2CO(g) + O2(g) → 2CO2 (g)
4. c) 2CO(g) + O2(g) → 2CO2 (g)
5. M) Neuhd rq;fpypj;njhlh; i`l;Nuh fhh;gd;fs;
5. b) linear chain hydrocarbons
6. m) Na+ madpapy; cl;fUtpd; ftu;r;rp tpir
6. a) the attractive force of nucleus is more in Na+ ion
Na mZit tpl mjpfk;.
than Na atom
7. M) ikl;Nlhfhz;l;hpahtpd; cl;gFjp(];l;Nuhkh)
7. b) mitochondrial matrix
8. m) nray; kpF flj;Jjy;
8. a) active transport
9. ,) m kw;Wk; M
9. c) both a and b
10. m) 9:3:3:1
10. a) 9:3:3:1
11. <) cly; gUkd; (Hint : kw;w 3-k; tif-1 ilahgb];
11. d) obese (Hint : Other 3 types are about Type-I diabetes mellitus) nky;ypl];)
12. d) scratch 12. <) scratch
PART –II gFjp – II

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13. Unit-1 : a) Both the assertion and the reason are true 13. myF-1: m) $w;W kw;Wk; fhuzk; ,uz;Lk;
and the reason is the correct explanation of the assertion. rup. NkYk; fhuzk; $w;Wf;F rupahd tpsf;fk;.
14. Unit-2 : TB Pg.no: 30 WTS Guide Pg.no:25 14. myF-2: Gj;jf vz;: 30 WTS g.vz; : 25
15. Unit-7 : TB Pg.no: 104 WTS Guide Pg.no: 100 15. myF-7: Gj;jf vz;: 102 WTS g.vz; : 102
16. Unit-10 : pH scale range is 0-14.where 0-7 is 16. myF-10: pH kjpg;G 0-14. 0-7 mkpyg;gz;G>
acid,7 is neutral and 7-14 is base. decreasing pH 7 eLepiy> 7-14 fhuj;jd;ik nfhz;lJ. pH kjpg;G
increase in acidic character. FiwAk; NghJ mkpyj;jd;ik mjpfupf;fpwJ.
pH of a solution is zero means, pH kjpg;G Rop vd;why;>
-log [H+]=0 -log [H+]=0
[H+]=1 [H+]=1
The concentration of hydrogen ion is 1. So it's i`l;u[d; madpapd; nrwpT 1. vdNt mjpf
highly acidic solution. mkpyj;jd;ik nfhz;ljhFk;.

17. Unit-12 :TB.Pg.no: 185 WTS Guide Pg.no: 154 17. myF-12: Gj;jf vz;: 182 WTS g.vz; : 160

18. Unit-13 : The sweat glands and sebaceous glands 18. myF-13: Kaypd; cly; ntg;gepiyia;
embedded in the skin regulate the body temperature. guhkupf;f tpau;it kw;Wk; vz;nza; Rug;gpfs;
cjTfpd;wd.
19. Unit-15: i)TB Pg.no: 227 WTS Guide Pg.no:188 19. myF-15: i)Gj;jf vz;: 224 WTS g.vz; : 194
ii) TB Pg.no: 227 WTS Guide Pg.no: 188 ii)Gj;jf vz;: 224 WTS g.vz; : 194

20. Unit-19 : TB Pg.no: 283 WTS Guide Pg.no: 234 20. myF-19: Gj;jf vz;: 283 WTS g.vz; : 248
21. Unit-20 : d) Both Assertion and reason is wrong. 21. myF-20: <)$w;W kw;Wk; fhuzk; ,uz;Lk; jtW

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10th – Science – PTA Question – 3: Answer Key
𝐿 𝐿
22. Unit-4 : Let length of single part be L’ = 22. myF-4: xU ghfj;jpd; ePsk;> L’ =
5 5
𝜌𝐿′ 𝜌𝐿 𝑅 5 𝜌𝐿′ 𝜌𝐿 𝑅 5
𝑅’ = = = = =1Ω 𝑅’ = = = = =1Ω
𝐴 5𝐴 5 5 𝐴 5𝐴 5 5
Effective resistance in Parallel connection, gf;f ,izg;gpy; njhFgad; kpd;jil>
1 1 1 1 1 1 5 1 1 1 1 1 1 5
= + + + + = =5Ω = + + + + = =5Ω
𝑅𝑝 𝑅’ 𝑅’ 𝑅’ 𝑅’ 𝑅’ 𝑅’ 𝑅𝑝 𝑅’ 𝑅’ 𝑅’ 𝑅’ 𝑅’ 𝑅’
1 1
Rp = = 0.2 Ω Rp = = 0.2 Ω
5 5

PART –III gFjp – III

23. Unit-1 : i) Weight of a body varies from one place to 23. myF-1: i) Gtp<u;g;G KLf;fkjpg;G Gtpapy; ,lj;jpw;F
another place on the Earth since it depends on the ,lk; khWgLtjhy;> vilapd; kjpg;Gk; ,lj;jpw;F ,lk;
acceleration due to gravity of the Earth (g) weight of a khWgLk;. nghUl;fspd; vil JUtg;gFjpapy;
body is more at the poles than at the equatorial region. mjpfkhfTk; > epyeLf; Nfhl; Lg; gFjpapy; FiwthfTk;
So the apples weigh more at poles than at equator. ,Uf; Fk; . vdNt epyeLf; N fhl; Lg; gFjpia tpl
JUtg;gFjpapy; Mg;gpspd; vil mjpfkhf ,Uf;Fk;.
Unit-2 : ii) TB Pg.no: 30 WTS Guide Pg.no: 25 myF-2:ii) Gj;jf vz;: 30 WTS g.vz; : 25

24. Unit-3 : i) Inside TB Pg.no: 39(Example – 2) 24. myF-3: i) Gj;jf cs;.g.vz; : 39 (v.fh – 2)

Unit-5 : ii) a) Infrasonic myF-5: ii)m) Fw;nwhyp

b) Application of Doppler effect M) lhg;sh; tpistpd; gad;ghL

25. Unit-6 : i) TB Pg.no: 89 WTS Guide Pg.no: 87 25. myF-6: i) Gj;jf vz;: 88 WTS g.vz; : 90
ii) Control rods : Used to control the number of ii) fl; Lg; gLj; J k; fopfs; : njhlu;tpidia
epiyepWj;jp epA+l;uhd;fspd; vz;zpf;ifiaf;
neutrons in order to have sustained chain reaction. fl;Lg;gLj;Jtjw;fhfg; gad;gLj;jg;gLfpwJ.
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Ex : Boron and cadmium rods. v.fh : Nghuhd; kw;Wk; fhl;kpak; fopfs;

26. Unit-8 : i) Dilute or concentrated nitric acid that 26. myF-8: i) mYkpaj;ij nray;glhepiyf;F
renders aluminium passive. Because, it does not cl; g Lj;
J tJ eP u ; j; j kw; W k; mlu; iel; upf; mkpyk; MFk;.
Vnddpy;> ,e;j mkpyk; mYkpdpaj;NjhL
attack aluminium, but it renders aluminium passive tpidGuptjpy;iy. khwhf mYkpdpaj;jpd; Nky; Mf;irL
due to the formation of an oxide film on its surface. glyk; cUthtjhy;> mjd; tpidgLk; jpwd;
jLf;fg;gLfpwJ.
Unit-10 : ii) TB Pg.no: 153 WTS Gui de Pg.no: 132
myF-10: ii)Gj;jf vz;: 151 WTS g.vz; : 137
27. Unit-7 : i) A is correct, R is wrong. 27. myF-7: i) A rup> R jtW
Unit-11:ii) TB Pg.no: 171 WTS Guide Pg.no: 142 myF-11: ii)Gj;jf vz;: 167 WTS g.vz; : 148

28. Unit-13: i)TB Pg.no: 199 WTS Guide Pg.no: 166 28.myF-13: i)Gj;jf vz;: 196 WTS g.vz; : 174

29.myF-15: i)Gj;jf vz;: 225 WTS g.vz; : 196

29. Unit-15 : i)TB Pg.no: 227 WTS Guide Pg.no: 190
myF-18: ii) F1 jiyKiw
Unit- 18: ii) F1 generation
T t
T t
t Tt tt
t Tt tt
t Tt tt
t Tt tt
[Pdhf;f tpfpjk; = 2Tt : 2tt = 2 : 2 = 1: 1
Genotypic ratio = 2Tt : 2tt = 2 : 2 = 1: 1

30. Unit-17: TB Pg.no: 258 WTS Guide Pg.no: 217 30.myF-17: Gj;jf vz;: 256 WTS g.vz; : 227

31. Unit-20 : i) TB Pg.no: 298 WTS Guide Pg.no: 246 31.myF-20: i) Gj;jf vz;: 297 WTS g.vz; :260
Unit-21:ii) TB Pg.no: 313 WTS Guide Pg.no:257 myF-21: ii) Gj;jf vz;: 313 WTS g.vz; :271

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10th – Science – PTA Question – 3: Answer Key
32. Unit-9: i) Mass of solute (sucrose) = 10g 32.myF-9:i)fiunghUspd; fdmsT = 10 fp
Mass Percentage of solution = 10% fiurypd; fdmsT = 10 %
Let, Mass of water = x ePhpd; epiw x vd;f
So, Mass of solution = x + 10 vdNt> fiurypd; epiw = x + 10
Mass of the sloute fiunghUspd; fdmsT
Mass % = × 100 fdmsT % = × 100
Mass of the solution fiurypd; fdmsT
10 10
10 = × 100 10 = × 100
x+10 x+10
x + 10 = 100 g x + 10 = 100 fp
Mass of water = 90 g ePhpd; epiw = 90 fp
Unit-10:ii) Inside TB Pg.no: 141 myF-10: ii) Gj;jf cs;.g.vz; : 139
PART –IV gFjp – IV
33A.Unit-1: i) TB Pg.no: 14 WTS Guide Pg.no: 11 33A. myF-1: i)Gj;jf vz;: 14 WTS g.vz; : 11
Unit-2: ii) Inside TB Pg.no: 27 myF-2: ii) Gj;jf cs;.g.vz; : 27

33B.Unit-5: i) Inside TB Pg.no: 60 33B.myF-5: i) Gj;jf cs;.g.vz; : 61

Unit-4: ii) Inside TB Pg.no: 53 (Figure – 4.12) myF-4: ii) Gj;jf cs;.g.vz; : 54 (glk; - 4.12)

34A.Unit-8: i) TB Pg.no: 123 WTS Guide Pg.no: 112 34A. myF-8: i)Gj;jf vz;: 120 WTS g.vz; : 116
Unit-11:ii)TB Pg.no: 171 WTS Guide Pg.no: 142 myF-11:ii) Gj;jf vz;: 167 WTS g.vz; : 148

34B.Unit-7:i)Mass of NH3 produced = 1000kg=106 g 34B. myF-7:i) ntspapLk; NH3 epiw =1000kg=106 g
molecular mass of NH3= 14 + (3 × 1) = 17 g NH3 -d; epiw = 14 + (3 × 1) = 17 g
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No of moles of NH3 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝐻3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓𝑁𝐻3
=
106
17
NH3 cs;s Nkhy;fspd; vz;zpf;if =
𝑁𝐻3 epiw 106
=
𝑁𝐻3 Nkhy;fspd; epiw 17
2 moles of NH3 is produced from 3 moles of H2
106 106 3 2 Nkhy; NH3 MdJ 3 Nkhy; H2ypUe;J ntsptUfpwJ.
∴ moles of NH3 is produced from × moles of H2 106 106 3
17 17 2
∴ Nkhy; NH3 MdJ × Nkhy; H2ypUe;J ntsptUfpwJ.
Required Mass of H2 = no of moles × molecular mass 17 17 2
106 3 Njitg;gLk; H2 tpd; epiw = Nkhy;fspd; vz;zpf;if
= × × (2 × 1) = 176.47 kg of H2 x %yf;$Wfspd; epiw
17 2
106 3
2 moles of NH3 is produced from 1 mole of N2 = × × (2 × 1) = 176.47 kg of H2
17 2
106 106 1 2 Nkhy; NH3 MdJ 1 Nkhy; N2ypUe;J ntsptUfpwJ.
∴ moles of NH3 is produced from × 2 moles of N2
17 17
106 106 1
106 1 28 ∴ Nkhy; NH3 MdJ × 2 Nkhy; H2ypUe;J ntsptUfpwJ.
Required Mass of N2= × × (14 × 2) = × 106 17 17
17 2 34
106 1 28
6 Njitg;gLk; N2 tpd; epiw = × 2 × (14 × 2) = 34 × 106
= 0.82353 × 10 g =823.53 kg of N2 17
= 0.82353 × 106 g =823.53 kg of N2
∴ Required mass of Nitrogen gas = 823.53 kg
∴ Njitg;gLk; H2 tpd; epiw =𝟏𝟕𝟔. 𝟒𝟕 fpfp
Required mass of Hydrogen gas = 𝟏𝟕𝟔. 𝟒𝟕 𝐤𝐠
Njitg;gLk; N2 tpd; epiw =𝟖𝟐𝟑. 𝟓𝟑 fpfp
Unit-10: ii) Faster Reaction - Digestion of food
slower Reaction -Rusting of iron myF-10:ii)Ntfkhf eilngWk; tpid - czT nrupj;jy;
nkJthf eilngWk; tpid - ,Uk;G JUg;gpbj;jy;
35A.Unit-14: i) TB Pg.no:215 WTS Guide Pg.no:182
35A. myF-14:i) Gj;jf vz;: 212 WTS g.vz; : 188
Unit-15:ii) Inside TB Pg.no: 220 myF-15: ii) Gj;jf cs;.g.vz; : 217
35B.Unit-16:i) Inside TB Pg.no: 232 35B. myF-16:i) Gj;jf cs;.g.vz; : 229
Unit-22:ii)TB Pg.no: 327 WTS GuidePg.no:267 myF-22:ii) Gj;jf vz;: 327 WTS g.vz; : 282

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10th – Science – PTA Question – 4: Answer Key
PTA Question - 4 Answer key
PTA – Model question Paper – 4 PTA – khjphp tpdhj;jhs; – 4
PART – I gFjp – I
1. d) both a and b 1. <) m kw;Wk; M
2. d) ohm metre 2. <) Xk; kPl;lh;
3. a) more than 83 3. m) 83[ tpl mjpfkhd
4. c) 25 ml ethanol in 75 ml of water 4. ,) 25 kpyp vj;jdhy; 75 kpyp ePhpy; cs;sJ
5. m) mjpfhpf;fpwJ
5. a) increases
6. ,) ii kl;Lk; rhp
6. c) only ii is correct
7. ,) H2O %yf;$Wfs; gpsf;fg;gLk;NghJ
7. c) when H2O is splitted
8. <) M kw;Wk; ,
8. d) both b and c
9. M) Mh;f;fpahg;nlhpf;];
9. b) archaeopterix 10. M) ,Ukak;
10. b) diploid 11. m) YA+f;Nfkpah
11. a) leukemia 12. m) notepad
12. a) notepad gFjp – II
PART –II 13. myF-2: Gj;jf vz;: 30 WTS g.vz; : 26
13. Unit-2 : TB Pg.no: 30 WTS Guide Pg.no: 25 14. myF-3: jtW. ntg;g Mw;wy; gupkhw;wj;jpd;
14. Unit-3 : False. Because During the process of NghJ Fiwe;j ntg;gepiyapy; cs;s nghUs;
transferring heat energy, the body at lower ntg;gg;gLj;jg;gLfpwJ. mjpf ntg;gepiyapy;
temperature is heated while the body at higher cs;s nghUs; Fspu;tpf;fg;gLfpwJ.

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temperature is cooled.
15. Unit-5 : Inside TB.Pg.no: 67 (problem – 1) 15. myF-5: Gj;jf cs;.g.vz; : 68 (jPu;T -1)
16. Unit-8 : TB Pg.no: 122 WTS Guide Pg.no: 111 16. myF-8: Gj;jf vz;: 119 WTS g.vz; : 114
17. Unit-13 : TB Pg.no: 198 WTS Guide Pg.no: 164 17. myF-13: Gj;jf vz;: 195 WTS g.vz; : 171
18. Unit-15 : TB Pg.no: 227 WTS Guide Pg.no: 189 18. myF-15: Gj;jf vz;: 224 WTS g.vz; : 195

19. Unit-18 : 19. myF-18:

[PNdhilg; gPNdhilg;
Genotype Phenotype capupdj;jpd; kugZ jfty;fs;
The hereditary information of The characters of an b.vd;.V tbtj;jpy; tho;ehs; capupdj;jpd; ntspapy;
KOtJk; mg;gbNa ,Uf;Fk;. njupAk; gz;Gfs; MFk;.
the organism in the form of organism which are
gene in the DNA and remains visible are known as mwptpay; Kiwfspd; %yk; capupdj;ij cw;W
the same throughout the life. phenotypes. jPu;khdpf;fyhk;. Nehf;Ftjd; %yk;
It can be determined by It can be determined by jPu;khdpf;fyhk;.
v.fh : ,uj;jtif> fz;zpd; v.fh : vil> clyikg;G>
scientific methods. observing the organism. epwk;> cauk;> kugZ Neha;fs; gwitfspd; myF
For eg., Blood group, eye colour, For eg., Weight, physique,
height, genetic diseases. beak of birds
20. myF-20: ,)$w;W kw;Wk; fhuzk; ,uz;Lk; rup.
20. Unit-20 : c) Both assertion and reason is correct 21. myF-22: Gj;jf vz;: 327 WTS g.vz; : 281
21. Unit-22 : TB Pg.no: 327 WTS Guide Pg.no: 266
22. myF-10: Kw = [H+] [OH-]
22. Unit-10 : Kw = [H+] [OH-] ePupd; madpg;ngUf;fk;,Kw = 1.00 × 10−14
ionic product of water,Kw = 1.00 × 10−14
ePupy; i`l;u[d; madpapd; nrwpT = 1.00 × 10−7
Conc. of hydrogen ions in water = 1.00 × 10−7
ePupy; i`l;uhf;rpy; madpapd; nrwpT = ?
Conc. of hydroxyl ions in water = ?
1.00 × 10−14 = [1.00 × 10−7 ] [OH-]
1.00 × 10−14 = [1.00 × 10−7 ] [OH-] 1.00 ×10−14
1.00 ×10−14 −7 [OH-] = = 1.00 × 10−7
-
[OH ] = = 1.00 × 10 1.00 ×10−7
1.00 ×10−7
ePupy; i`l;uhf;rpy; madpapd; nrwpT = 1.00 × 10−7
Conc. of hydroxyl ions in water = 1.00 × 10−7

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10th – Science – PTA Question – 4: Answer Key
PART –III gFjp – III
23. Unit-1:i) a) Moment of a couple 23. myF-1: i) m) ,ul;ilfspd; jpUg;Gj;jpwd;
b) Unlike parallel forces M) vjpnujph; jpirapy; nray;gLk; rkkw;w ,iztpirfs;
1 1
Unit-2:ii) Power of a lens (P) = = -2D myF-2: ii) nyd;rpd; jpwd; (P) = = -2D
𝑓 𝑓
1 1
f= = -0.5m f= = -0.5m
−2 −2

24. Unit- 5: i) i = 90̊ – 50̊ = 40̊ 24. myF-5: i) i = 90̊ – 50̊ = 40̊
angle of incident sound (i)= angle of reflected sound (r) gLNfhzk;(i) =vjpnuhspg;Gf;Nfhzk;; (r)
Therefore, Angle of Reflection of sound (r) = 40̊ vdNt xypapd; vjpnuhspg;Gf; Nfhzk; = 40̊
Unit-6 : ii) Nuclear Fission reaction : myF-6: ii) mZf;fU gpsT tpid :

X – 56Ba141, Y – 36Kr92 X – 56Ba141, Y – 36Kr92

25. Unit-7 : i) a) Neutrons b) atomicity 25. myF-7: i)m) epA+l;uhd;fs; M) mZf;fl;L vz;

Unit-8 : ii) TB Pg.no: 122 WTS Guide Pg.no: 111 myF-8: ii) Gj;jf vz;: 119 WTS g.vz; :115

26. Unit-9 : i) TB Pg.no: 135 WTS Guide Pg.no: 122 26. myF-9: i)Gj;jf vz;: 133 WTS g.vz; : 125
Unit-11 : ii) a) A & R are correct, R explains the A myF-11:ii)m) A kw;Wk; R rup> R MdJ AI tpsf;FfpwJ.

27. Unit-7 : i) mass % of carbon = 27.28 = mass of C 27.myF-7: i)fhu;gdpd; epiw%=27.28=fhu;gdpd; epiw
mass % of oxygen = 72.73 = mass of O Mf;rp[dpd; epiw % = 72.73 = Mf;rp[dpd; epiw
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑟𝑏𝑜𝑛 fhu;gdpd; epiw
no of moles of C = fhu;gd; Nkhy;fspd; vz;zpf;if =
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=
𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑟𝑏𝑜𝑛
27.28
12
= 2.27 ≅ 2 =
27.28
12
fhu;gd; mZtpd; epiw

= 2.27 ≅ 2
72.73 72.73
no of moles of O = = 4.54 ≅ 4 Mf;rp[d; Nkhy;fspd; vz;zpf;if = = 4.54 ≅ 4
16
16
molecular formula - C2 O4 (or) 2 CO2 %yf;$W tha;g;ghL - C2 O4 (my;yJ) 2 CO2
molecular mass =(2 × 12) + (4 × 16) = 𝟖𝟖𝒈 %yf;$W epiw =(2 × 12) + (4 × 16) = 𝟖𝟖 fp
myF-8: ii) fhg;gu;igiul;Lfs; fhg;gupd; Kf;fpa
Unit-8 :ii) Copper pyrites is the prime ore of copper.
It is concentrated by froth floatation method. jhJ MFk;. ,J Eiukpjg;G Kiwapy;
mlu;gg
; pf;fg;gLfpwJ. fhuzk; : Nyrhd
Reason : lighter ores, such as sulphide ores, are jhJf;fshd> ry;igL jhJf;fs;> ,k;Kiwapy;
concentrated by this method. mlu;gg
; pf;fg;gLfpd;wd. v.fh : fhg;gu; igiul;];
Ex : Copper pyrites (CuFeS2), Zinc blende(Zns) (CuFeS2)> [pq;f; gpsd;l; (Zns)
28. Unit-13: i) TB Pg.no: 198 WTS Guide Pg.no: 165 28.myF-13: i)Gj;jf vz;: 195 WTS g.vz; : 172
Unit-15 :ii) TB Pg.no: 227 WTS Guide Pg.no: 189 myF-15: ii)Gj;jf vz;: 224 WTS g.vz; : 196

29. Unit-16 : TB Pg.no: 240 WTS Guide Pg.no: 201 29.myF-16: Gj;jf vz;: 238 WTS g.vz; : 210

30. Unit-17: i) TB Pg.no: 258 WTS Guide Pg.no: 216 30.myF-17: i)Gj;jf vz;: 255 WTS g.vz; : 225
Unit-20 :ii) TB Pg.no: 298 WTS Guide Pg.no: 245 myF-20: ii)Gj;jf vz;: 297 WTS g.vz; : 259
31. Unit-21 :i)TB Pg.no: 312 WTS Guide Pg.no: 256 31.myF-21: Gj;jf vz;: 313 WTS g.vz; : 270
Unit-22 :ii) TB Pg.no: 327 WTS Guide Pg.no: 266 myF-22: ii)Gj;jf vz;: 327 WTS g.vz; : 281
32. Unit-1: i) The power lines which are made up of
32.myF-1:i) cNyhfq;fshy; Md kpd; fk;gpfs;
metals expands on heating during hot summer, Nfhilf;fhyq;fspy; ntg;gkhjypd; fhuzkhf
this makes the lines to hang low. tpuptiltjhy; jho;thfj; njhq;Ffpd;wd.
Unit-4: ii) Inside TB Pg.no: 51(Problem – 6) myF-4: ii) Gj;jf cs;.g.vz; : 51 (jPu;T – 6)
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10th – Science – PTA Question – 4: Answer Key
PART –IV gFjp – IV
33A. Unit-1: TB Pg.no: 14 WTS Guide Pg.no:13 33A. myF-1: Gj;jf vz;: 14 WTS g.vz; : 13

33B.Unit-4: i) TB Pg.no: 57 WTS Guide Pg.no:55 33B.myF-4: i) Gj;jf vz;: 57 WTS g.vz; : 55
Unit-6: ii) Inside TB Pg.no:83 myF-6: ii) Gj;jf cs;.g.vz; : 82

34A. myF-7: i)Gj;jf vz;: 103 WTS g.vz; : 104

34A.Unit-7: i) TB Pg.no: 104 WTS Guide Pg.no: 102
myF-9: ii) Gj;jf vz;: 132 WTS g.vz; : 123
Unit-9:ii)TB Pg.no: 135 WTS Guide Pg.no: 119

34B. myF-7:i) m) CaCO3(s)+Heat → CaO(s) + CO2(g)

34B.Unit-7:i)a) CaCO3(s)+Heat → CaO(s) + CO2(g) in ntg;gr;rpijT tpidfspy; gpizg;Gfis
thermal decomposition reaction, heat is supplied to cilg;gjw;F ntg;gk; jug;gLfpwJ. ,J xU
break the bonds. It is a irreversible reaction. kPshtpid MFk;.
CaCO3(s)+Heat → CaO(s) + CO2(g) in thermal CaCO3(s)+Heat → CaO(s) + CO2(g) in thermal
deomposition reaction, heat is supplied to break the ntg;gr;rpijT tpidfspy; gpizg;Gfis
bonds. And if the reaction is carried out in a closed cilg;gjw;F ntg;gk; jug;gLfpwJ. ,e;j tpid
xU %ba fydpy; eilg;ngw;why; mJ
vessel, it reaches a chemical equilibrium.
Ntjpr;rkepiyia milAk;.
b) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) M) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
This reaction is a neutralization reaction. Nkw;fz;l tpid eLepiyahf;fy; tpidahFk;.
Reason : reaction of sodium hydroxide with fhuzk; : Nrhbak; i`l;uhf;irL kw;Wk; i`l;Nuh
hydrochloric acid is a typical neutralization FNshupf; mkpyj;jpw;F ,ilNaahd tpid
reaction. Here, sodium replaces hydrogen from eLepiyahf;fy; tpidahFk;. ,q;F Nrhbak;>
i`l;u[id i`l;NuhFNshupf; mkpyj;jpypUe;J
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hydrochloric acid forming sodium chloride, a
neutral soluble salt.
,lg;ngau;r;rp nra;fpwJ. ,jd; tpisthf Nrhbak;
FNshiuL vd;w eLepiyahd ePupy; fiuAk; cg;G
fpilf;fpwJ.
Unit-11:ii) a) Ethanol, b) Ethanoic acid myF-11: ii) m)vj;jdhy;> M) vj;jdhapf; mkpyk;
c) sodium salt of ethanoic acid ,) vj;jdhapf; mkpyj;jpd; Nrhbak; cg;G

35A.Unit-12: i) Inside TB Pg.no: 175(Figure – 12.1)

35A. myF-12: i) Gj;jf cs;.g.vz; : 172 (glk;- 12.1)
Unit-14:ii)a) Phloem transports food. When,
myF-14:ii)m) GNshak; czit flj;JfpwJ.
leaves synthesize food, it transports food from ,iyfspy; czT cw;gj;jpahFk; NghJ
source to sink. On the basis of plants need (or) Njhw;WthapypUe;J Njf;fplj;jpw;F flj;jg;gLfpwJ.
season, transfer of food is reversed from sink to jhtuj;jpd; Njitf;Nfw;gTk; fhykhw;wj;jpw;F Vw;gTk;
source. Therefore the movement of food in czT Njf;fplj;jpypUe;J Njhw;Wthapw;F flj;jg;gLfpwJ.
phloem can be in all direction. vdNt> GNshaj;jpd; topahf czTnghUshdJ
midj;J gFjpfSf;Fk; gy jpirfspy; flj;jg;gLfpwJ.
b) Inside TB Pg.no:205 (‘Do you know’ content) M) Gj;jf cs;.g.vz; : 202 (cq;fSf;F njupAkh)

35B.Unit-18:i) TB Pg.no:272 WTS GuidePg.no:225 35B. myF-18: i) Gj;jf vz; :271 WTS g.vz; : 237
Unit-19:ii)Inside Pg.no:284 WTS GuidePg.no:237 myF-19: ii)Gj;jf vz;: 283 WTS g.vz; : 251
Unit-21:iii)Inside Pg.no:312 WTS GuidePg.no:256 myF-21: ii)Gj;jf vz;: 313 WTS g.vz; : 270

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10th – Science – PTA Question – 5: Answer Key
PTA Question - 5 Answer key
PTA – Model question Paper – 5 PTA – khjphp tpdhj;jhs; – 5
PART – I gFjp – I
𝐹 nghUj;jkhd tpiliaj; Njh;e;njLj;J vOJf.
1. c)
4 𝐹
2. d) all the above 1. ,)
4
3. a) thermal Expansion 2. <) ,it midj;Jk;
4. a) ionic 3. m) ntg;gtphpT
4. m) madpj;jd;ik
5. a) strong affinity to water
5. m) ePhpd; kPJ mjpf ehl;lk;
6. b) remain same 6. M) xNu khjphpahf ,Uf;fpd;wd
7. a) sanguivorous 7. m) rhq;fpNthu];
8. c) corpus callosum 8. ,) fhh;g]; fNyhrk;
9. c) the centre 9. ,) ikaj;jpy;
10. d) repetitive 10. <) kPz;Lk; kPz;Lk; tUk; njhlh;
11. c) ii and iii 11. ,) ii kw;Wk; iii
12. ,) tpz;Nlh];> ypdf;];
12. c)Windows, LINUX
gFjp – II
PART –II 13. myF-3:M) $w;W kw;Wk; fhuzk; Mfpa ,uz;Lk;
13. Unit-3 : b) Both the assertion and the reason are true but rup. Mdhy;> fhuzk; $w;Wf;F rupahd
the reason is not the correct explanation of the assertion tpsf;fky;y.
14. Unit-4 : SI unit of consumption of electrical energy is 14. myF-4: kpd;dhw;wy; Efu;tpd; SI myF thl;
watt second. Its larger unit is kilowatt hour (kWh). tpehb. ,jd; ngupa myF fpNyh thl; kzp(kWh).
xU fpNyh thl; kzp vd;gjid xU A+dpl;
One kilowatt hour also known as one unit of electrical kpd;dhw;wy; MFk;. xU fpNyh thl; kzp vd;gJ
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energy. One kilowatt hour means that an electric
power of 1000 watt has been utilized for an hour.
1000 thl; kpd;rhuk;
Efug;gl;Ls;sjhFk;.
xU kzpNeuj;jpy;

15. Unit-9 : TB Pg.no: 135 WTS Guide Pg.no: 120 15. myF-9: Gj;jf vz;: 132 WTS g.vz; : 124
16. Unit-10 : TB Pg.no: 154 WTS Guide Pg.no: 134 16. myF-10: Gj;jf vz;: 151 WTS g.vz; : 139
17. Unit-12 : Glycolysis is the common step in aerobic 17. myF-12: fhw;W Rthrj;jpw;Fk; fhw;wpy;yh
and anaerobic pathway. Glycolysis takes place in Rthrj;jpw;Fk; nghJthd epfo;r;rp fpisf;fhyp]p];.
cytoplasm of the cell. ,J irl;Nlhgpshrj;jpy; eilngWfpwJ.

18. Unit-14 : TB Pg.no: 215 WTS Guide Pg.no: 178 18. myF-14: Gj;jf vz;: 212 WTS g.vz; : 184
19. Unit-16 : TB Pg.no: 240 WTS Guide Pg.no: 201 19. myF-16: Gj;jf vz;: 237 WTS g.vz; : 209
20. Unit-19 : TB Pg.no: 283 WTS Guide Pg.no: 234 20. myF-19: Gj;jf vz;: 282 WTS g.vz; : 248
21. Unit-21 : TB Pg.no: 312 WTS Guide Pg.no: 255 21. myF-21: Gj;jf vz;: 313 WTS g.vz; : 269
22. Unit-5 : TB Pg.no: 71 WTS Guide Pg.no: 73 22. myF-5: Gj;jf vz;: 71 WTS g.vz; : 75
PART –III gFjp – III
23.Unit-1:i) F=5N, a = 5 cm s-2 = 0.05 m s-2 23. myF-1: i) F=5N, a = 5 cm s-2 = 0.05 m s-2
𝐹 5 𝐹 5
F = ma⇒ m = = = 100 kg F = ma⇒ m =𝑎 = 0.05 = 100 kg
𝑎 0.05
m = 100 kg m = 100 kg
Unit-2:ii) h = 3cm ; u = 10 cm ; v = 20 cm myF-2: ii) h = 3cm ; u = 10 cm ; v = 20 cm
𝑣 20 𝑣 20
Magnification m = = =2 cUg;ngUf;fk; m = = =2
𝑢 10 𝑢 10
ℎ′ ℎ′
Magnification m = cUg;ngUf;fk; m =
ℎ ℎ
ℎ′ = 2 × 3 = 6cm ℎ′ = 2 × 3 = 6cm
ℎ′ = 6 cm ℎ′ = 6 cm

24. Unit-4 : TB Pg.no: 56 WTS Guide Pg.no: 53 24. myF-4: Gj;jf vz;: 57 WTS g.vz; :53

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10th – Science – PTA Question – 5: Answer Key
25. Unit-6 : i) mass number = 226 – 4 = 222 25. myF-:6 i) epiw vz; = 226 – 4 = 222
atomic number = 88 - 2 = 86 mZ vz; = 88 - 2 = 86
Unit-6 : ii) Mass defect in the reaction (m) = 1 kg myF-6: ii) tpidapd;NghJ epiwtO (m) = 1 fpfp
Velocity of light (c) = 3 × 108 m s-1 xspapd; jpirNtfk; (c) = 3 × 108 m s-1
By Einstein’s equation, [d;];Bd; epiw Mw;wy; rkd;ghL E = mc2
Energy released E = mc2
vdNt> E =1× (3 × 108)2 = 9 × 1016 J (or) 0.9 × 1017 J
So, E = 1× (3 × 108)2 = 9 × 1016 J (or) 0.9 × 1017 J

26. Unit-7 : i) TB Pg.no: 104 WTS Guide Pg.no: 101 26. myF-7: i)Gj;jf vz;: 103 WTS g.vz; : 103
myF-8: ii)M) A rup> R jtW
Unit-8 : ii) b) A is correct, R is Wrong.

27.Unit-11:i)ii)iii) TB Pg.no:171WTS Guide Pg.no: 145 27.myF-11: i)ii)iii)Gj;jf vz;:168 WTS g.vz; : 151

28. Unit-12: i) TB Pg.no: 185 WTS Guide Pg.no: 155 28.myF-12: i)Gj;jf vz;: 182 WTS g.vz; : 162
Unit-15 :ii) TB Pg.no: 227 WTS Guide Pg.no: 189 myF-15: ii)Gj;jf vz;: 225 WTS g.vz; : 196

29. Unit-17 : TB Pg.no: 258 WTS Guide Pg.no: 217 29.myF-17: Gj;jf vz;: 256 WTS g.vz; : 227

30. Unit-18: i) TB Pg.no: 272 WTS Guide Pg.no: 225 30.myF-18: i)Gj;jf vz;: 271 WTS g.vz; : 237
Unit-19:ii)a),b)TB Pg.no:283WTS Guide Pg.no: 233 myF-19: ii)Gj;jf vz;: 282 WTS g.vz; : 247

31. Unit-20 :i)TB Pg.no: 298 WTS Guide Pg.no: 247 31.myF-20: Gj;jf vz;: 297 WTS g.vz; : 260

32. Unit-9: i) Inside TB Pg.no: 128 32.myF-9: i) Gj;jf cs;.g.vz; : 125

Impact of Temperature : Solubility of CO2 in water ntg;gepiyapd; jhf;fk; : ePupd; ntg;gepiyia
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decreases with the increase in temperature.
Impact of Pressure : Solubility of CO2 in water
mjpfupf;Fk; NghJ CO2 tpd; fiujpwd; FiwfpwJ.
mOj;jj;jpd; jhf;fk; : ePupd; mOj;jj;ij mjpfupf;Fk;
increases with the increase in Pressure. NghJ CO2 tpd; fiujpwd; mjpfupf;fpwJ.
ii) Pure water does not conduct electricity. This is because ii)J}a ePH kpd;rhuk; flj;Jtjpy;iy. ,jpy; madpfs;
of the absence of ions in it . It can conduct electricity , ,y;yhjNj ,jw;Ff; fhuzk;. ehk; xU rpl;bif cg;ig
when we add a pinch of salt to it. NrHf;Fk;NghJ mJ kpd;rhuj;ij flj;j KbAk;.

PART –IV gFjp – IV

33A. Unit-1: i) WTS Guide Pg.no:12(Newton’s 2 law) 33A. myF-1:i)WTS g.vz; :12 (epA+l;ldpd; 2k; tpjp)
nd

Unit-4: ii) Voltmeter is used to measure the potential myF-4: ii) kpd;dOj;j NtWghl;bid mstpLk;
difference. It is connect in parallel in a circuit. fUtp Nthy;l; kPl;lu;. kpd;Rw;wpy; ,jid
gf;f,izg;gpy; ,izf;fNtz;Lk;.
33B.Unit-5: i) Inside TB Pg.no:68 (Problem-5) 33B.myF-5: i) Gj;jf cs;.g.vz; : 69(jPu;T – 5)
ii) Wavelength (𝜆) =3000 Å = 3000 × 10−10 𝑚 ii) miyePsk; (𝜆) 3000 Å = 3000 × 10−10 kP
velocity of light (𝑐 ) = 3 × 108 ms-1 xypapd; jpirNtfk; (𝑐) = 3 × 108 ms-1
𝐶 𝐶
frequency(𝜈) = mjph;ntz(; 𝜈)=
𝜆 λ;
3×108 3×108 3×108 3×108
= = = =
3000×10−10 3×10−7 3000×10−10 3×10−7
frequency(𝜈) = 1015 Hz mjph;ntz(; 𝜈)= 1015 Hz
Unit-6: iii)TB Pg.no: 89 WTS Guide Pg.no: 89 myF-6: iii) Gj;jf vz;: 88 WTS g.vz; : 91

34A.Unit-7: i) TB Pg.no: 103 WTS Guide Pg.no: 101 34A. myF-7: i)Gj;jf vz;: 103 WTS g.vz; : 103
Unit-10:ii) Inside TB Pg.no: 146 myF-10: ii) Gj;jf cs;.g.vz; : 144

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10th – Science – PTA Question – 5: Answer Key
Mass of the solute fiunghUspd; epiw
34B.Unit-9:i) Solubility = Mass of the solvent
× 100 34B. myF-9:i) fiujpwd; = × 100
fiug;ghdpd;; epiw
10 10
= × 100 = 20 g = × 100 = 20 fp
50 50
Unit-11:ii) Inside TB Pg.no: 168 (Figure – 11.3) myF-11: ii) Gj;jf cs;.g.vz; : 164 (glk;-11.3)

35A.Unit-13:i) TB Pg.no:198 WTS GuidePg.no:165 35A. myF-13: i) Gj;jf vz;: 195 WTS g.vz; : 172
Unit-14:ii) TB Pg.no:215 WTS GuidePg.no: 178 myF-14:ii) Gj;jf vz;: 212 WTS g.vz; : 184

35B.Unit-21:i) TB Pg.no:313 WTS GuidePg.no:259 35B. myF-21: i) Gj;jf vz;: 314 WTS g.vz; : 273
Unit-22:ii) Inside TB Pg.no: 322 myF-22: ii) Gj;jf cs;.g.vz; : 322

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10th – Science – PTA Question – 6: Answer Key
PTA Question - 6 Answer key
PTA – Model question Paper – 6 PTA – khjphp tpdhj;jhs; – 6
PART – I gFjp – I
1. d) 1 dyne (F=ma=1 gcms-2 = 1dyne) 1. <) 1 ild; (F=ma=1 gcms-2 = 1 ild;)
2. b) an ideal gas 2. M) ey;ypay;G thA
3. b) 20 kHz 3. M) 20 kHz
4. c) homo atomic molecule 4. ,) xj;j mZ %yf;$W
5. c) A-ii, B-iii, C-iv, D-i 5. ,) A-ii, B-iii, C-iv, D-i
6. v) CH4 < CH2 = CH2 < CH ≡ CH 6. ,) CH4 < CH2 = CH2 < CH ≡ CH
7. a) root hair 7. m) Ntu;j;J}tp
8. b) Insulin 8. M) ,d;Rypd;
9. d) Charles Darwin 9. <) rhu;y]; lhu;tpd;
10. c) Carcinoma 10. ,) fhu;rpNdhkh
11. a) Tidal energy 11. m) Xj Mw;wy;
12. c) file 12. ,) Nfhg;G
PART –II gFjp – II
13. Unit-2 : 13. myF-2:

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14. Unit-3 : TB Pg.no: 38 (Problem-1) 14. myF-3: Gj;jf vz;: 39 (jPu;T – 1)

15. Unit-6 : TB Pg.no: 88 WTS Guide Pg.no: 82 15. myF-6: Gj;jf vz;: 87 WTS g.vz; : 88

16. Unit-10 : TB Pg.no: 154 WTS Guide Pg.no: 133 16. myF-10: Gj;jf vz;: 151 WTS g.vz; : 138

17. Unit-13 : TB Pg.no: 198 WTS Guide Pg.no: 164 17. myF-13: Gj;jf vz;: 195 WTS g.vz; : 171
18. Unit-15 : Inside TB Pg.no: 223 18. myF-15: Inside TB Pg.no: 220
19. Unit-17 : Inside Pg.no: 258 WTS Guide Pg.no:215 19. myF-17: Gj;jf vz;: 255 WTS g.vz; : 225
20. Unit-20 : TB Pg.no: 298 WTS Guide Pg.no: 245 20. myF-20: Gj;jf vz;: 297 WTS g.vz; : 259
21. Unit-22 : TB Pg.no: 327 WTS Guide Pg.no: 266 21. myF-22: Gj;jf vz;: 327 WTS g.vz; : 281

22. Unit-8 : i) Hydrogen element belongs to 1st group. 22. myF-8: i) i`l;u[d;- 1-tJ njhFjp jdpkk;.
rfg;gpizg;G Mu kjpg;G 0.37 A° nfhz;l jdpkk;.
and its covalent radius value is 0.37𝐴°. Chemical
i`l;u[dpd; NtjpFwpaPL- H,
symbol of Hydrogen is H and hydrogen molecule is H2. i`l;u[d; %yf;$wpd; FwpaPL - H . 2

ii) Aluminium is a metal belongs to boron family. ii) Nghuhd; FLk;gj;ij Nru;e;jJ.
It reduces iron oxide into iron. ,Uk;G Mf;iril ,Uk;ghf xLf;FfpwJ.
𝐹𝑒2 𝑂3 + 2𝐴𝑙 → 2 𝐹𝑒 + 𝐴𝑙2 𝑂3 +Heat 𝐹𝑒2 𝑂3 + 2𝐴𝑙 → 2 𝐹𝑒 + 𝐴𝑙2 𝑂3 + ntg;g Mw;wy;
mYkpdpak;> rikay; ghj;jpuq;fs; nra;tjw;F
Aluminium is used to make household utensils. gad;gLfpwJ.

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10th – Science – PTA Question – 6: Answer Key
PART –III gFjp – III
23. i) Unit-3 : 𝐾 = (𝐹 + 460) × 5⁄9 23. i) myF-3: 𝐾 = (𝐹 + 460) × 5⁄9
= (80 + 460) × 5⁄9 = 300 K = (80 + 460) × 5⁄9 = 300 𝐾
ii) Unit-4: TB Pg.no: 57 WTS Guide Pg.no: 57 ii myF-4: Gj;jf vz;: 58 WTS g.vz; : 58
24. Unit-5 : i) Inside Pg.no: 71 WTS Guide Pg.no: 71 24. myF-5: i) Gj;jf vz;: 71 WTS g.vz; : 73
Unit-6 : ii) a) Hydrogen bomb myF-6: ii) m) i`l;u[d; Fz;L
b) effective functioning of heart M) ,jaj;ij rPuhf nray;gl
25. Unit-9 : i) TB Pg.no: 135 WTS Guide Pg.no: 123 25. myF-9: i) Gj;jf vz;: 133 WTS g.vz; : 127
ii) a) Unit-10 : TB Pg.no: 153 WTS Guide Pg.no: 129 ii) m) myF-10 : Gj;jf vz;: 150 WTS g.vz; : 134
b) Unit-10 : TB Pg.no: 146 M) myF-10 : Gj;jf vz;: 144
Rate of the chemical reaction increases on raising the ntg;gepiyia caHj;Jk;NghJ Ntjptpidapd;
temperature. Because adding heat to the reactants Ntfk; ; mjpfupf;fpwJ. Vnddpy;
tpidgLnghUs;fSld; ntg;gj;ij
provides energy to break more bonds and thus speed mjpfupg;gjhy; mjpf gpizg;Gfis cilf;Fk;
up the reaction. Mw;wiy toq;FfpwJ> ,jdhy; Ntjptpidapd;
Ntfk; mjpfupf;fpwJ.
26. Unit-11 : i) TB Pg.no: 171 WTS Guide Pg.no: 144 26. myF-11: i)Gj;jf vz;: 167 WTS g.vz; : 150
Unit-10 : ii) TB Pg.no: 154 WTS Guide Pg.no: 133 myF-10: ii)Gj;jf vz;: 151 WTS g.vz; : 138
27.myF-11: Nru;kk; A - vj;jdhy; epwkw;w jputk;
27. Unit-11 : kw;Wk; vupRit nfhz;lJ.
Compound A – Ethanol having Colourless liquid & vj;jdhypd; Mtpia 573K ntg;gepiyapy;>
burning taste R+Nlw;wg;gl;l jhkpuj;jpd; kPJ i`l;u[d; ePf;fk;
Vapour of Ethanol is passed over heated copper eilngw;W mrpl;lhy;bi`L cUthfpwJ.
573K, it is dehydrogenated to acetaldehyde
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jhkpuk; tpidA+f;fpahf gad;gLfpwJ.
Copper is used as catalyst.
28. myF-12 : Gj;jf vz;: 173 (glk; - 12.2)
28. Unit-12 : TB Pg.no: 176 (Figure – 12.2)
29. i)myF-13: Gj;jf vz;: 195 WTS g.vz; : 170
29. i) Unit-13 : TB Pg.no: 198 WTS Guide Pg.no: 163 ii) myF-14: Nyz;l;];Bdu; kw;Wk; tPdu; Rh
ii)Unit-14 : Rh factor was discovered by Landsteiner fhuzpiaf; fz;lwpe;jdu;. uPr]; ,df;Fuq;fpid
and Wiener. It is named after Rhesus monkey. (Rhesus monkey) gad;gLj;jpajhy; Rh fhuzp vd;W
miof;fg;gLfpwJ.
30. Unit-17: i) TB Pg.no: 258 WTS Guide Pg.no: 216 30.myF-17: i)Gj;jf vz;: 255 WTS g.vz; : 225
Unit-20 :ii) TB Pg.no: 298 WTS Guide Pg.no: 245 myF-20: ii)Gj;jf vz;: 297 WTS g.vz; : 259

31. Unit-19 : TB Pg.no: 283 WTS Guide Pg.no: 235 31.myF-19: Gj;jf vz;: 283 WTS g.vz; : 249

32. Unit-1: i) Inside Pg.no: 12 (Problem – 3) 32.myF-1: i) Gj;jf vz; : 12 (jPu;T – 3)

Unit-2: ii) Least distance of distinct vision of a human myF-2: i)nghJthf kdpjf; fz;zpd; njspTW
is 25 cm. fhl;rpapd; kPr;rpWj; njhiyT kjpg;G 25 nr.kP
PART –IV gFjp – IV
33A. Unit-2: i) TB Pg.no: 30 WTS Guide Pg.no: 26 33A. myF-2: i) Gj;jf vz;: 30 WTS g.vz; : 27
Unit-4: myF-4 :
ii) a) Total resistance of the circuit=20 Ω + 4 Ω ii) m) kpd;Rw;wpd; nkhj;j kpd;jil=20 Ω + 4 Ω
= 24 Ω = 24 Ω
𝑉 6
𝑉 6
b) The current through the circuit = 𝑅 = 24 = 0.25 𝐴 M) kpd;Rw;wpy; ghAk; kpd;Ndhl;lk; = 𝑅 = 24 = 0.25 𝐴
c) The potential difference across the resistance ,) kpd;jilahf;fpapd; FWf;Nf kpd;dOj;j
= 0.25 x 4 = 1 V NtWghL = 0.25 x 4 = 1 V
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10th – Science – PTA Question – 6: Answer Key
33B.Unit-3:i) TB Pg.no: 40 WTS Guide Pg.no:40 33B. myF-3: i) Gj;jf vz;: 41 WTS g.vz; : 41
Unit-5:ii) TB Pg.no: 71 WTS Guide Pg.no:72 ii) myF-5: Gj;jf vz;: 71 WTS g.vz; : 74
Unit-6:iii) TB Pg.no: 82 iii) myF-6: Gj;jf vz;: 81

34A. Unit-7: i) TB Pg.no: 104 WTS Guide Pg.no: 101 34A. myF-7: i) Gj;jf vz;: 103 WTS g.vz; : 103
Unit-11: ii) TB Pg.no: 171 WTS Guide Pg.no: 144 myF-11: ii) Gj;jf vz;: 167 WTS g.vz; : 150
34B.Unit-8: i)
34B. myF-8: i) m)
a)Top to bottom, atomic radius increases. m) NkypUe;J> fPohf mZ Muk; mjpfupf;fpwJ.
Reason: Due to increase in valence shell number. fhuzk; : ntspf;$l;L vz; mjpfupg;gJ.
Left to Right, atomic radius decreases. ,lkpUe;J tykhf mZ Muk; FiwfpwJ.
Reason : valence shell number remains same, fhuzk; : ntspf;$l;L vz; khwhJ> vdpDk;>
but number of protons increases, increasing GNuhl;lhd;fspd; vz;zpf;if mjpfupg;gjhy;>
the attraction of protons over electrons. Thus, GNuhl;lhd; kw;Wk; vyf;l;uhd;fSf;F
atomic radius shrinks. ,ilNaahd <u; gG; tpir mjpfupj; J> mZtpd;
cUtsT RUq;FfpwJ.
b) Down the group, electron affinity decreases.
M) NkypUe;J> fPohf vyf;l;uhd; ehl;lk; FiwfpwJ.
Reason : Atomic radius increases and so, fhuzk; : mZ Muk; mjpfupg;gjhy;>
valence electrons are loosely bound. Therefore ntspf;$l;L vyf;l;uhd;fs; ,yFthf
electron affinity decreases. gpizf;fg;gl;bUf;fpd;wd. vdNt vyf;l;uhd;
Left to Right, electron affinity increases. ehl;lk; FiwfpwJ.
Reason : Atomic radius decreases and so, ,lkpUe;J tykhf vyf;l;uhd; ehl;lk; mjpfupf;fpwJ.
electron affinity increases. fhuzk; : mZ Muk; Fiwtjhy; vyf;l;uhd;
ehl;lk; mjpfupf;fpwJ.

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c) Left to right, ionisation energy increases.
Reason : As atomic radius decreases, more
,) ,lkpUe;J tykhf madpahf;Fk; Mw;wy; mjpfupf;fpwJ.
fhuzk; : ,lkpUe;J tykhf nry;Yk; NghJ
energy is required to remove the electrons as mZ Muk; Fiwtjhy;> vyf;l;uhd;fis
we go from left to right. ePf;Ftjw;F mjpf Mw;wy; Njitg;gLfpwJ.
Down the group, ionisation energy decreases. NkypUe;J> fPohf madpahf;Fk; Mw;wy; FiwfpwJ.
Reason : Atomic radius increases and so, fhuzk; : mZ Muk; mjpfupg;gjhy;>
valence electrons are loosely bound. Less ntspf;$l;L vyf;l;uhd;fs; ,yFthf
energy is required to remove the electrons as gpizf;fg;gl;bUf;fpd;wd. vdNt
vyf;l;uhd;fis ePf;Ftjw;F Fiwthd Mw;wy;
we go down the group. Njitg;gLfpwJ.

Unit-10: ii) Manganese dioxide acts as a catalyst and myF-10: ii) khq;fdPR il Mf;irL
increases the reaction rate of the given reaction. tpidA+f;fpahf nray;gLfpwJ. NkYk;> tpidapd;
Ntfj;ij mjpfupf;fpwJ.
35A.Unit-16:i) Parthenocarpic fruits – The seedless 35A. myF-16: i) fUTwhf;fdpfs; - Mf;rpd;fisj;
fruits developed without fertilization that are njspg;gjhy; fUTWjy; eilngwhkNyNa
induced by external application of auxins. tpijapyhf; fdpfs; cUthjy; J}z;lg;gLfpwJ.
E.g. Watermelon, Tomato. ,k;Kiwapy; cUthFk; fdpfs; fUTwhf;fdpfs;
vdg;gLk;. v.fh : ju;G+rzp> jf;fhsp
Unit-15:ii)TB Pg.no: 228 WTS Guide Pg.no: 192 myF-15: ii) Gj;jf vz;: 225 WTS g.vz; : 199

35B.Unit-18:i) TB Pg.no: 272 WTS Guide Pg.no: 225 35B. myF-18: i) Gj;jf vz;: 271 WTS g.vz; : 238
Unit-21:ii) TB Pg.no: 313 WTS Guide Pg.no: 258 myF-21: ii) Gj;jf vz;: 313 WTS g.vz; : 272

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