part–a : chemistry · 2 1. tc ,flfvd ,flm dk 0.2 g csathu ds 20 g esa feyk;k tkrk gs rks csathu...
TRANSCRIPT
2
1. tc ,flfVd ,flM dk 0.2 g csathu ds 20 g esa feyk;k
tkrk gS rks csathu dk fgekad 0.45°C ls de gks tkrk gSA
;fn ,flfVd ,flM casthu esa laxqf.kr gksdj Mkbej (fr;)
cukrk gS rks ,flfVd ,flM dk çfr'krrk laxq.ku gksxk
(csathu ds fy, Kf = 5.12 K kg mol–1)
(1) 94.6%
(2) 64.6%
(3) 80.4%
(4) 74.6%
mÙkj mÙkj mÙkj mÙkj mÙkj (1)
gy%gy%gy%gy%gy% 0.45 = i(5 .12) 0.2 / 60
100020
×
⇒ i = 0.527
( )3 3 21–
2
2CH COOH CH COOH
α α
⇒ i 1–2
α=
⇒ 0.527 = 1–2
α
⇒2
α = 0.473
⇒ α = 0.946
∴ % laxq.ku = 94.6%
2. CoCl3.6H
2O ds 0.1 M foy;u ds 100 mL dks AgNO
3
ds vkf/D; esa vfHkÑr djus ij 1.2 × 1022 vk;u voksfirgksrs gSaA ladqy gS%
(1) [Co(H2O)
5Cl]Cl
2.H
2O
(2) [Co(H2O)
4Cl
2]Cl.2H
2O
(3) [Co(H2O)
3Cl
3].3H
2O
(4) [Co(H2O)
6]Cl
3
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% AgNO3 ds fefy eksy =
22
23
1.2 101000 20
6 10
× × =×
CoCl3.6H
2O ds fefy eksy = 0.1 × 100 = 10
∴ CoCl3.6H
2O ds çR;sd eksy ls nks DyksjkbM vk;u çkIr
gksrs gSaA
∴ [Co(H2O)
5Cl]Cl
2.H
2O
3. eksuksukbVªs'ku vfHkfØ;k esa fuEu esa ls dkSu lk ;kSfxdesVk mRikn dh egRoiw.kZ ek=kk mRiUu djsxk\
(1)
NHCOCH3
(2)
OH
(3)
OCOCH3
(4)
NH2
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy%
NH2
H+
NH3
NO2
NH3
NO2
NH3
NO2
+
NH3
+
NO2
(51%)
(47%)
(2%)
4. tc Dyksjhu xSl BaMs ,oa ruq tyh; NaOH ds lkFkvfHkfØ;k djrh gS rks çkIr gksus okys mRikn gksaxs%
(1) Cl– rFkk –
2ClO (2) ClO– rFkk –
3ClO
(3) –
2ClO rFkk –
3ClO (4) Cl– rFkk ClO–
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy% 2 2Cl 2NaOH NaCl NaOCl H O+ ⎯⎯→ + +
B.Mk o ruq lkfs M; egkbiksDyksjkbV
5. fod.kZ lEcU/ ds dkj.k] yhfFk;e rFkk eSXuhf'k;e nksuksa dbZ,d tSls xq.k çnf'kZr djrs gSa fiQj Hkh] og ,d tks xyrgS] gS%
(1) yhfFk;e rFkk eSXuhf'k;e] nksuksa ds gh ukbVªsV xje djusij NO
2 rFkk O
2 nsrs gSa
(2) nksuksa kkjh; dkcksZusV cukrs gSa
(3) nksuksa ?kqyu'khy ckbdkcksZusV cukrs gSa
(4) nksuksa ukbVªkbM cukrs gSa
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy% Mg kkjh; dkcksZusV] ( )23 23MgCO .Mg OH .3H O cukrk
gS ysfdu bl çdkj ds kkjh; dkcksZusV Li kjk ugha cuk,tkrsA
PART–A : CHEMISTRY
JEE Main Paper I Code C (In Hindi)
3
6. ,d ty çfrn'kZ esa ih- ih- ,e- (ppm) Lrj dh fuEu½.kk;uksa dh lkUnzrk gSA
F– = 10; 2–
4SO 100= ; 2–
3NO 50=
og@os ½.kk;u tks ty çfrn'kZ dks ihus ds fy, vuqi;qDrcukrk gS@cukrs gSa] gS@gSa%
(1) ek=k 2–
4SO
(2) ek=k –
3NO
(3) 2–
4SO rFkk –
3NO nksuksa
(4) ek=k F–
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy% is; ty esa F– dh vuqes; lhek 1 ppm rd gksrh gSAF– dh 10 ppm ls vf/d lkUnzrk ds dkj.k gfîó;ksa dk k;gks tkrk gSA
7. fuEu cgqydksa esa ls dkSu ls cgqyd esa ty vi?kVuvfHkfØ;k lfUufgr gS\
(1) Vsjhyhu (2) ukbykWu 6
(3) csosQykbV (4) ukbykWu 6, 6
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy% osQizksysDVe ty vi?kfVr gksdj osQizksbd vEy nsrk gS tksla?kuu ij mRikn ukbykWu-6 nsrk gSA
O
NH H O3
+
O
C
HO (CH ) – NH2 5 2
(osQizksysDVe) (osQizksbd vEy)
8. fVUMy çHkko rHkh fn[kk;h iM+sxk tc fuEu 'krZ larq"Vgksrh gS%
(a) ifjksfir d.kksa dk O;kl] ç;qDr çdk'k ds rjaxnSè;Zdh rqyuk esa cgqr NksVk gksA
(b) ifjksfir d.kksa dk O;kl] ç;qDr çdk'k ds rjaxnSè;Zdh rqyuk esa cgqr NksVk ugha gksA
(c) ifjksfir çkoLFkk rFkk ifjksi.k ekè;e ds viorZukadifjek.k yxHkx ,d tSls gksaA
(d) ifjksfir çkoLFkk rFkk ifjksi.k ekè;e ds viorZukadifjek.k cgqr fHkUu gksaA
(1) (b) rFkk (c) (2) (a) rFkk (d)
(3) (b) rFkk (d) (4) (a) rFkk (c)
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy% fV.My çHkko ds fy, ijhfkIr çkoLFkk rFkk ifjksi.k ekè;edk viorZukad i;kZIr :i ls fHkUu gksus pkfg, vkSj ijhfkIrizkoLFkk ds d.kksa dk vkdkj ç;qDr rjaxnSè;Z ls T;knk fHkUuugha gksuk pkfg,A
9. ,d nqcZy vEy (HA) dk pKa rFkk ,d nqcZy kkjd
(BOH) dk pKb Øe'k% 3.2 rFkk 3.4 gSaA muds yo.k (AB)
ds foy;u dk pH gksxk%
(1) 1.0 (2) 7.2
(3) 6.9 (4) 7.0
mÙkj mÙkj mÙkj mÙkj mÙkj (3)
gy%gy%gy%gy%gy% pH = ( )a b
17 + pK – pK
2
= ( )17 + 3.2 – 3.4
2
= 6.9
10. fuEu vfHkfØ;k esa izkIr eq[; mRikn gS%
O
O
DIBAL-H
COOH
(1) CHO
CHO
(2) CHO
COOH
OH
(3) CHO
CHO
OH
(4)
COOH
CHO
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy% DIBAL-H ,LVj rFkk dkcksZfDlfyd vEyksa dks ,fYMgkbM esavipf;r dj nsrk gS
OO
DIBAL-H
OHO
H
CHOCOOH
JEE Main Paper I Code C (In Hindi)
4
11. ,d tyh; KOH foy;u esa fuEu esa ls dkSu lk ;kSfxd,d vipk;d 'koZQjk osQ :i esa O;ogkj djsxk\
(1) HOH C2
CH OCH2 3
OH
OH
OH
O
(2) OHOH C2
CH OH2
OCOCH3
HO
OH
(3) OHOH C2
CH OH2
HO
OH
(4) OHOH C2
CH OH2
OCH3
HO
OH
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy% 'koZQjk,a ftuesa eqDr ,ukejh –OH lewg gksrs gSa os vipk;d'kdZjk,a gksrh gSaA
O CH – OH2
O – C– CH3
HO
OH
OH
CH2
O KOH (aq)
O CH – OH2
HO
OH
OH
CH2
OH + CH COOK3
eqDr ,ukejh lewg
12. fuEu :ikUrj.k osQ fy, vfHkdeZdksa dk lgh Øe gksxk%
O
CHO
HO
CH3
CH3
CH3HO
(1) [Ag(NH3)2]+OH–, CH
3MgBr, H+/CH
3OH
(2) [Ag(NH3)2]+OH–, H+/CH
3OH, CH
3MgBr
(3) CH3MgBr, H+/CH
3OH, [Ag(NH
3)2]+OH–
(4) CH3MgBr, [Ag(NH
3)2]+OH–, H+/CH
3OH
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy%
O
C = OH
[Ag(NH ) ] OH3 2
+ –
O
C = OHO
CH – OH/H3
+
,LVjhdj.k
O
C – OCH3
O
(i) CH MgBr
(3 )
3
eksy(ii) H O
2
HO
HO – CCH
3
CH3
CH3
13. fuEu esa ls dkSulh Lih'kht vuqpqEcdh; ugha gS\(1) B
2(2) NO
(3) CO (4) O2
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy% CO esa 14 bysDVªkWu (le) gS ∴ ;g çfrpqEcdh; gS
NO esa 15 bysDVªkWu (fo"ke) gS ∴ ;g vuqpqEcdh; gS rFkkblds π∗2p vk.kfod dkd esa ,d v;qfXer bysDVªkWu gSA
B2 esa 10 bysDVªkWu (le) gS ysfdu fiQj Hkh ;g vuqpqEcdh;
gS rFkk blds π2px ,oa π2p
y (s-p feJ.k) esa nks v;qfXer
bysDVªkWu gSA
O2 esa 16 bysDVªkWu (le) gSA ysfdu fiQj Hkh ;g
vuqpqEcdh; gS rFkk blds π*2px ,oa π*2p
y vk.kfod dkd
esa nks v;qfXer bysDVªkWu gSA
14. fuEu esa ls dkSu] tert-BuONa ds lkFk vfHkoQr djus rFkkczksehu ty osQ feykus ij] czksehu osQ jax dks jaxghu djusesa vleFkZ gksrk gS\
(1)
O
Br
(2)
O
Br
(3)Br
C H6 5
(4)
O
Br
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy%Br
CH3
C
CH3
CH3
Na+
O–
O
CH3
C
CH3
CH3
O
( )mRikn
O
mijksDr mRikn esa dksbZ C=C ;k C≡C ca/ ugha gS] vr%;g Br
2 ty ijhk.k ugha nsxkA
JEE Main Paper I Code C (In Hindi)
5
15. fuEu esa ls dkSu lh vfHkfØ;k vip;ksip; (fjMkWDl)vfHkfØ;k dk mnkgj.k gS\
(1) XeF6 + 2H
2O → XeO
2F
2 + 4HF
(2) XeF4 + O
2F
2 → XeF
6 + O
2
(3) XeF2 + PF
5 → [XeF]+ –
6PF
(4) XeF6 + H
2O → XeOF
4 + 2HF
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy% Xe; +4 (XeF4 esa) ls +6 (XeF
6 esa) vkWDlhÑr gks tkrk gSA
vkWDlhtu +1 (O2F
2 esa) ls 'kwU; (O
2 esa) vipf;r gksrh gSA
16. ΔU ftlds cjkcj gS] og gS%
(1) lerkih dk;Z (2) le&vk;rfud dk;Z
(3) lenkch dk;Z (4) :¼ks"e dk;Z
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy% :¼ks"e çØe ds fy,] q = 0
∴∴∴∴∴ Å"ekxfrdh ds çFke fu;e ds vuqlkj
ΔU = W
17. fuEu esa ls dkSu lk v.kq vuqukfnd :i ls U;wure fLFkj gS\
(1)
O
(2)
(3)
O
(4)N
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% gkykafd fodYiksa esa fn;s x;s lHkh v.kq vuquknh fLFkj gS ysfdufodYi (2) esa fn;k x;k ;kSfxd U;wure vuquknh fLFkj gS(vU; rhuksa ,jksesfVd gS)A
O O–
18. SN1 vfHkfØ;k osQ fy, fuEu gSykbMksa dh vfHkfØ;kRedrk
dk c<+rk Øe gS%
3 2 3
(I)
CH CHCH CH|Cl
3 2 2
(II)
CH CH CH Cl
3 6 4 2(III)
p – H CO – C H – CH Cl
(1) (II) < (III) < (I) (2) (III) < (II) < (I)
(3) (II) < (I) < (III) (4) (I) < (III) < (II)
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy% SN1 vfHkfØ;k dk osx α dkcZ/uk;u dk LFkkf;Ro
I. CH – CH CH CH3 2 3
– –
Cl
CH – CH CH CH3 2 3
– –
II. CH – CH CH Cl3 2 2
– – CH – CH CH3 2 2
–
III.
CH2 – Cl
OCH3
CH2
OCH3
blfy,] II < I < III
vr% ;g dkcZ/uk;u dk c<+rk LFkkf;Ro gS rFkk bl dkj.kgSykbMksa dh vfHkfØ;kRedrk blh Øe esa c<+rh gSA
19. ,d dkcksZusV (M2CO
3) ds 1 xzke dks HCl ds vkf/D; esa
vfHkfØr fd;k tkrk gS vkSj mlls 0.01186 eksy CO2 iSnk
gksrh gSA M2CO
3 dk eksyj æO;eku g mol–1 esa gS%
(1) 11.86 (2) 1186
(3) 84.3 (4) 118.6
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy% M2CO
3 + 2HCl → 2MCl + H
2O + CO
2
2 3 2M CO CO
n n=
2 3M CO
10.01186
M=
2 3M CO
M = 1
0.01186 = 84.3 g/mol
20. ,d dkcZfud vEy dk lksfM;e yo.k 'X' lkUnz H2SO
4 ds
lkFk cqncqnkgV nsrk gSA 'X' vEyh; tyh; CaCl2 ds lkFk
vfHkfØ;k djrk gS vkSj lisQn voksi nsrk gS tks KMnO4
ds vEyh; foy;u dks jaxghu cuk nsrk gSA 'X' gS%
(1) Na2C
2O
4(2) C
6H
5COONa
(3) HCOONa (4) CH3COONa
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% 2 2 4 2 4 2 4 2 2 4(X)
Na C O H SO Na SO H C O+ ⎯⎯→ +lkUnz vkDW lfs yd vEy
2 4
2
H SO
2 2 4 2
–H O
H C O CO COΔ⎯⎯⎯⎯⎯⎯→ ↑ + ↑
cqncqnkgV
lkUnz
2 2 4 2 2 4(X)
Na C O CaCl CaC O 2NaCl+ ⎯⎯→ ↓ +'osr voksi
– 2– 2
4 2 4 2 22MnO 5C O 16H 2Mn 10CO 8H O
+ ++ + → + +
JEE Main Paper I Code C (In Hindi)
6
21. ,d LoLFk euq"; ds 'kjhj esa ek=kk dh nf"V ls cgqrk;r ls feyusokys rRo gS% vkWDlhtu (61.4%); dkcZu (22.9%), gkbMªkstu(10.0%); rFkk ukbVªkstu (2.6%)A 75 kg otu okys ,dO;fDr ds 'kjhj ls lHkh 1H ijek.kqvksa dks 2H ijek.kqvksa ls cnyfn;k tk; rks mlds Hkkj esa tks o`f¼ gksxh] og gS%
(1) 10 kg (2) 15 kg
(3) 37.5 kg (4) 7.5 kg
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy% gkbMªkstu dk nzO;eku = 10
75 7.5 kg100
× =
1H dks 2H kjk çfrLFkkfir djus ij 7.5 kg dk çfrLFkkiu15 kg kjk gksxkA
∴ usV o`f¼ = 7.5 kg
22. fuEu vfHkfØ;k esa izkIr gksus okyk eq[; mRikn gS%
C H6 5
C H6 5
(+)
t-BuOK
Δ
BrH
(1) (–)C6H
5CH(OtBu)CH
2C
6H
5
(2) (±)C6H
5CH(OtBu)CH
2C
6H
5
(3) C6H
5CH=CHC
6H
5
(4) (+)C6H
5CH(OtBu)CH
2C
6H
5
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gy%gy%gy%gy%gy%
BrH
C H6 5
C H6 5
t-BuOK
Δ(E-2) C H
6 5
C H6 5
(+)
23. fn;k x;k gS]
C(graphite)
+ O2(g) → CO
2(g);
ΔrH° = – 393.5 kJ mol–1
H2(g) +
1
2O
2(g) → H
2O(I);
ΔrH° = –285.8 kJ mol–1
CO2(g) + 2H
2O(l) → CH
4(g) + 2O
2(g);
ΔrHº = +890.3 kJ mol–1
Åij fn;s x;s Å"ejklk;fud lehdj.kksa ds vk/kj ij298 K ij vfHkfØ;k
C(graphite)
+ 2H2(g) → CH
4(g) ds Δ
rH° dk eku gksxk:
(1) –144.0 kJ mol–1
(2) +74.8 kJ mol–1
(3) +144.0 kJ mol–1
(4) –74.8 kJ mol–1
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gy%gy%gy%gy%gy% C(graphite)
+ O2(g) → CO
2(g);
ΔrH° = –393.5 kJ mol–1 ...(i)
2 2
1H (g) O (g)
2+ → H
2O(l);
ΔrH° = –285.8 kJ mol–1 ...(ii)
CO2(g) + 2H
2O(l) → CH
4(g) + 2O
2(g);
ΔrH° = 890.3 kJ mol–1 ...(iii)
bu izfØ;kvksa dks ykxw djus ij]
(i) + 2 × (ii) + (iii) gesa izkIr gksrk gS
C(graphite)
+ 2H2(g) → CH
4(g);
ΔrH° = –393.5 –285.8 × 2 + 890.3
= –74.8 kJ mol–1
24. fuEu vfHkfØ;kvksa esa] ZnO Øe'k% dk;Z djsxk%
(a) ZnO + Na2O → Na
2ZnO
2
(b) ZnO + CO2 → ZnCO
3
(1) vEy rFkk kkjd
(2) kkjd rFkk vEy
(3) kkjd rFkk kkjd
(4) vEy rFkk vEy
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% (a) esa] Na2O kkjh; vkWDlkbM gksus ds dkj.k ZnO vEyh;
vkWDlkbM ds :i esa dk;Z djrk gSA
(b) esa] ZnO kkjh; vkWDlkbM ds :i esa dk;Z djrk gS D;ksafdCO
2 vEyh; vkWDlkbM gSA
25. gkbMªkstu ijek.kq ds frh; cksj dkk dk v¼ZO;kl gksxk %
(IySad fLFkjkad h = 6.6262 × 10–34 Js;
bysDVªkWu dk æO;eku = 9.1091 × 10–31 kg;
bysDVªkWu dk vkos'k e = 1.60210 × 10–19 C;
fuokZr dk ijkoS|qrkad ∈0 = 8.854185 × 10–12 kg –1m–3A2
(1) 2.12 Å
(2) 1.65 Å
(3) 4.76 Å
(4) 0.529 Å
mÙkj mÙkj mÙkj mÙkj mÙkj (1)
gy%gy%gy%gy%gy% r = 2
0
na = 0.529 ×4
Z
= 2.12 Å
JEE Main Paper I Code C (In Hindi)
7
26. nks vfHkfØ;kvksa] R1 rFkk R
2 ds iwoZ pj?kkrkadh xq.kd ,d
tSls gSaA R1 dh lafØ;.k ÅtkZ R
2 osQ lafØ;.k ÅtkZ ls
10 kJ mol–1 T;knk gSA ;fn vfHkfØ;k R1 rFkk R
2 ds fy,
300 K ij nj fu;rk ad Øe'k% k1 rFkk k
2 gk s a rk s
ln(k2/k
1) fuEu esa ls fdlds cjkcj gksxk\
(R = 8.314 J mol–1K–1)
(1) 4 (2) 8
(3) 12 (4) 6
mÙkj mÙkj mÙkj mÙkj mÙkj (1)
gy%gy%gy%gy%gy% k1 = 1
a–E /RT
Ae
k2 = 2
a–E /RT
Ae
2
1
k
k =
( )1 2a a
1E – E
RTe
1 2a a2
1
E – Ekln =
k RT
= 3
10 104
8.314 300
× ≈×
27. ,d /krq iQyd dsfUær ?ku lajpuk esa fØLVfyr gksrh gSA;fn blds ,dd lsy dh dksj yEckbZ 'a' gS] rks èkfRodfØLVy esa nks ijek.kqvksa ds chp lfÂdVre nwjh gksxh%
(1)a
2(2) 2a
(3) 2 2 a (4) 2 a
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% FCC ds fy, ,d iQyd fuEu çdkj dh gksrh gS
A
BC
a
Δ ABC kjk
2a2 = 16r2
⇒ 2 21r a
8=
⇒1
r a2 2
=
lfUudVre çfrosf'k;ksa dh nwjh = 2r = a
2
28. og xzqi ftlesa lebysDVªkWuh Lih'kht gSa] gS
(1) O–, F–, Na+, Mg2+
(2) O2–, F–, Na+, Mg2+
(3) O–, F–, Na, Mg+
(4) O2–, F–, Na, Mg2+
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gy%gy%gy%gy%gy% Mg2+, Na+, O2– ,oa F– izR;sd esa 10 bysDVªkWu gSA
29. fn;k x;k gS]
–
2
°
Cl /ClE 1.36V,=
3
°
Cr /CrE –0.74V+ =
2– 3+
2 7
°
Cr O /CrE 1.33 V,= – 2+
4
°
MnO /MnE 1.51 V.=
fuEu esa ls çcyre vipk;d gS %
(1) Cl– (2) Cr
(3) Mn2+ (4) Cr3+
mÙkj mÙkj mÙkj mÙkj mÙkj (2)
gy%gy%gy%gy%gy% Cr3+ ds fy,, 3+ 2–
2 7Cr /Cr O°E = – 1.33 V
Cl– ds fy,, –
2Cl /Cl
E° = – 1.36 V
Cr ds fy,, 3Cr/Cr
E° + = 0.74 V
2Mn
+ ds fy,] 2 –
4Mn /MnO
E° + = – 1.51 V
Cr ds fy, E° /ukRed gS vr% ;g çcy vipk;d gS
30. ijkDlkbM dh mifLFkfr esa] 3-esfFky&isUV&2&bZu] HBr osQlkFk vfHkfØ;k djus ij ,d ladyu mRikn cukrk gSAmRikn osQ fy, lEHko f=kfoe leko;fo;ksa dh la[;k gksxh%
(1) pkj (2) N%
(3) 'kwU; (4) nks
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy%gy%gy%gy%gy% CH3 – CH = C – CH – CH
2 3
CH3
HBr
R O2 2
3- -2-esfFky isUV bZu
CH3 – CH – C – CH – CH
2 3
CH3
Br H
mRikn (X)
* *
D;ksafd mRikn (X) esa nks fdjSy osQUnz gSa rFkk ;g vlefergS] vr% blesa oqQy f=kfoe leko;o = 22 = 4.
JEE Main Paper I Code C (In Hindi)
8
31. lekdy
3
4
4
1 cos
π
π +∫dx
x cjkcj gS %
(1) 4 (2) –1
(4) 2 (3) –2
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy
3 3
4 42
2
4 4
1sec
2 22cos
2
dx xdx dx
x
π π
π π=∫ ∫
3
4
4
tan1 2
12
2
x
π
π
⎡ ⎤⎢ ⎥
= ⎢ ⎥⎢ ⎥⎣ ⎦
3tan tan
8 8
π π= −
1 cos2 1 2 14
tan8 12 1
1 cos4
⎡ π−⎢ π − −⎢ = = =
π +⎢ +⎢⎣
31 cos
3 2 14tan 2 1
38 2 11 cos
4
⎤π− ⎥π +⎥= = = +
π − ⎥+⎥⎦
( 2 1) ( 2 1)
2
= + − −
=
32. ekuk tan ,( 1)n
nI xdx n= >∫ gSA ;fn
5 5
4 6 tanI I a x bx C+ = + + gS] tgk¡ C ,d lekdyu
vpj gS] rks Øfer ;qXe (a, b) cjkcj gS %
(1)1, 1
5
⎛ ⎞−⎜ ⎟⎝ ⎠
(2)1,0
5
⎛ ⎞−⎜ ⎟⎝ ⎠
(3)1,1
5
⎛ ⎞−⎜ ⎟⎝ ⎠
(4)1,0
5
⎛ ⎞⎜ ⎟⎝ ⎠
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy tan , 1n
nI xdx n= >∫
4 6
4 6
4 2
(tan tan )
tan sec
I I x x dx
x xdx
+ = +
=
∫
∫
ekuk tanx = t
sec2x dx = dt
4
5
5
5
1tan
5
t dt
tC
x C
=
= +
= +
∫
1, 0
5a b= =
33. ks=k (x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4y rFkk 1≤ +y x
dk ks=kiQy (oxZ bdkb;ksa) esa gS %
(1)7
3
(2)5
2
(3)59
12
(4)3
2
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy
y
x′x
′y
(1, 2
)
(2, 1)
O
x = 0
xy
+ = 3
x = 2x = 1
(0, 1)
Nk;kafdr Hkkx dk ks=kiQy1 22 2
0 1
1 (3 )4 4
x xx dx x dx
⎛ ⎞ ⎛ ⎞= + − + − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∫ ∫
5 sq. unit
2=
PART–B : MATHEMATICS
JEE Main Paper I Code C (In Hindi)
9
34. ,d cDls esa 15 gjh rFkk 10 ihyh xsansa gSaA ;fn ,d&,ddjds ;kn`PN;k] izfrLFkkiuk lfgr] 10 xsansa fudkyh tk,¡]rks gjh xsanksa dh la[;k dk izlj.k gS %
(1) 4
(2)6
25
(3)12
5
(4) 6
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy n = 10
p(gjh xsan fudkys tkus dh izkf;drk) = 15
25
3 2,
5 5p q∴ = =
izlj.k(X) = n.p.q
= 6 12
1025 5
⋅ =
35. ;fn ( ) ( )2 sin 1 cos 0+ + + =dyx y x
dx rFkk y(0) = 1 gS]
rks 2
π⎛ ⎞⎜ ⎟⎝ ⎠
y cjkcj gS %
(1)1
3−
(2)4
3
(3)1
3
(4)2
3−
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy (2 sin ) ( 1)cos 0+ + + =dyx y x
dx
(0) 1, ?2
π⎛ ⎞= =⎜ ⎟⎝ ⎠
y y
1 cos0
1 2 sin+ =
+ +x
dy dxy x
ln| 1| ln(2 sin ) ln+ + + =y x C
( 1)(2 sin )+ + =y x C
x = 0, y = 1 j[kus ij
(1 1) 2+ ⋅ = C ⇒ C = 4
vc, ( 1)(2 sin ) 4+ + =y x
2
π=x ds fy,
( 1)(2 1) 4+ + =y
41
3+ =y
4 11
3 3= − =y
36. ekuk ω ,d lfEeJ la[;k ,slh gS fd 2ω + 1 = z tgk¡
3z = − gSA ;fn
2 2
2 7
1 1 1
1 1 3
1
k−ω − ω =ω ω
gS
rks k cjkcj gS%
(1) –1
(2) 1
(3) –z
(4) z
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy 2ω + 1 = z , 3z i=
1 3
2
i− +ω = → bdkbZ dk ?kuewyA
C1 → C
1 + C
2 + C
3
2 2 2 2
2 7 2 2
1 1 1 1 1 1 3 1 1
1 1 1 0
1 1 0
− − ω ω = ω ω = ω ω
ω ω ω ω ω ω
= 3 (ω2 – ω4)
= 1 3 1 3
32 2
i i⎡ ⎤⎛ ⎞ ⎛ ⎞− − − +−⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
= 3 3i−
= –3z
∴ k = –z
37. ekuk ˆ ˆ ˆ ˆ ˆ2 2a i j k b i j= + − = +
rFkk gSA ekuk c
,d ,slk
lfn'k gS fd | | 3, ( ) 3c a a b c− = × × =
rFkk c
vkSj
a b×
ds chp dks.k 30° gS] rks a c⋅
cjkcj gS %
(1) 5 (2)1
8
(3)25
8(4) 2
JEE Main Paper I Code C (In Hindi)
10
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy | ( ) | 3a b c× × =
ˆ ˆ ˆ2 2a b i j k× = − +
| | | | sin 30 3a b c⇒ × ° =
| | 3a a b= = ×
| | 2c⇒ =
| | 3c a− =
2 2| | | | 2( ) 9c a a c⇒ + − ⋅ =
9 3 22
2a c
− −⋅ = =
38. U;wure ks=kiQy okys ,sls o`Ùk] tks oØ y = 4 – x2 rFkkjs[kkvksa y = |x| dks Li'kZ djrk gS] dh f=kT;k gS %
(1) ( )4 2 1−
(2) ( )4 2 1+
(3) ( )2 2 1+
(4) ( )2 2 1−
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy dsUnz : (0, 4 – r)
∴ x – y = 0 dh (0, 4 – r) l s yEc dh yEckb Z
0 4
2
− += rr gS
⇒ 4 2− = ±r r
4
1 2
=±
r , 4
1 2
≠−
r
∴ 4( 2 1)= −r
39. ;fn 1
0,4
x⎛ ⎞∈⎜ ⎟⎝ ⎠ ds fy, 1
3
6tan
1 9
x x
x
− ⎛ ⎞⎜ ⎟⎝ ⎠−
dk vodyu
. ( )x g x gS] rks g(x) cjkcj gS %
(1) 3
3
1 9
x
x−(2) 3
3
1 9x+
(3) 3
9
1 9x+(4)
3
3
1 9
x x
x−mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy 1( ) 2 tan (3 )f x x x−=
10,
4x
⎛ ⎞∈⎜ ⎟⎝ ⎠ ds fy,
3
9( )
1 9
xf x
x
′ =+
3
9( )
1 9g x
x=
+
40. ;fn leqPp; 0, 1, 2, 3, ....., 10 eas ls nks fofHk la[;k,¡fudkyh xb±] rks muds ;ksxiQy rFkk muds varj ds fujiskeku] nksuksa ds pkj ds xq.kd gksus dh izkf;drk gS %
(1)14
45
(2)7
55
(3)6
55
(4)12
55
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy rjhdksa dh dqy la[;k = 11C2= 55
vuqdwy rjhds fuEu gSa(0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10)
izkf;drk= 6
55
41. ( )32
cot coslim
2x
x x
xπ→
−π −
cjkcj gS %
(1)1
8(2)
1
4
(3)1
24(4)
1
16
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy3
2
cot coslim
( 2 )π→
−π −x
x x
x
2
π − =x t j[kus ij
30
tan sinlim
8→
−t
t t
t
=
2
30
sin 2sin2
lim8t
tt
t→
⋅=
1
16.
42. ( ) ( )21 10 21 10
1 1 2 2C C C C− + − +
( ) ( )21 10 21 10
3 3 4 4.....C C C C− + − + +
( )21 10
10 10C C− dk eku gS %
(1) 220 – 29 (2) 220 – 210
(3) 221 – 211 (4) 221 – 210
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy 21 21 21 21 21 21
1 2 10 0 1 21
1... ... 1
2C C C C C C+ + + = + + + −
= 220 – 1
( )10 10 10 10
1 2 10... 2 1+ + + = −C C C
∴ visfkr ;ksxiQy = (220 – 1) – (210 – 1) = 220 – 210
JEE Main Paper I Code C (In Hindi)
11
43. rhu ?kVukvksa A, B rFkk C ds fy,
P(A vFkok B eas ls dsoy ,d ?kfVr gksrh gS)
= P(B vFkok C eas ls dsoy ,d ?kfVr gksrh gS)
= P(C vFkok A eas ls dsoy ,d ?kfVr gksrh gS) 1
4= rFkk
P(lHkh rhu ?kVuk,¡ ,d lkFk ?kfVr gksrh gaS) 1
16= gS]
rks izkf;drk fd de ls de ,d ?kVuk ?kfVr gks] gS %
(1)7
64
(2)3
16
(3)7
32
(4)7
16
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy1
( ) ( ) ( )4
P A P B P A B+ − ∩ =
1( ) ( ) ( )
4P B P C P B C+ − ∩ =
1( ) ( ) ( )
4P C P A P A C+ − ∩ =
( ) ( ) ( ) ( ) ( )
3( )
8
P A P B P C P A B P B C
P A C
+ + − ∩ − ∩
− ∩ =
1( )
16P A B C∩ ∩ =∵
∴3 1 7
( )8 16 16
P A B C∪ ∪ = + =
44. ekuk ,d ÅèokZ/j ehukj AB ,slh gS fd mldk fljk A
Hkwfe ij gSA ekuk AB dk eè; fcanq C gS rFkk Hkwfe ijfLFkr fcanq P ,slk gS fd AP = 2AB ;fn ∠BPC = β gS]rks tan β cjkcj gS %
(1)2
9
(2)4
9
(3)6
7
(4)1
4
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gy gy gy gy gy
B
C
A
x
x
4xP
βθ
1tan
4θ =
( ) 1tan
2θ + β =
∴
1tan
14
1 21 tan
4
+ β=
− β
gy djus ij 2
tan9
β =
45. ,d nh?kZo`Ùk] ftldk dsUnz ewy fcUnq ij gS] dh mRdsUnzrk
1
2 gSA ;fn mldh ,d fu;rk x = – 4 gS] rks mlds fcanq
31,
2
⎛ ⎞⎜ ⎟⎝ ⎠
ij mlds vfHkyac dk lehdj.k gS %
(1) 4x + 2y = 7
(2) x + 2y = 4
(3) 2y – x = 2
(4) 4x – 2y = 1
(4)
gygygygygy
x = –4
1
2=e
4− = −a
e
4− = − ×a e
2=a
vc, 2 2 2(1 ) 3b a e= − =
nh?kZoÙk dk lehdj.k
2 2
14 3
+ =x y
JEE Main Paper I Code C (In Hindi)
12
vfHkyEc dk lehdj.k
3
1 2
1 3
4 2 3
−− =
×
yx
⇒ 4 2 1 0− − =x y
46. ;fn fdlh /uiw.kk±d n ds fy,] f?kkrh lehdj.k
x(x + 1) + (x + 1) (x + 2) + .....
( 1)( ) 10x n x n n+ + − + =
ds nks Øfed iw.kk±dh; gy gS] rks n cjkcj gS %
(1) 10 (2) 11
(3) 12 (4) 9
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy lehdj.k dks iqu%O;ofLFkr djus ij
2 1 3 5 .... (2 1)nx n x+ + + + + −
1 2 2 3 ... ( 1) 10n n n+ ⋅ + ⋅ + + − =
2 2 ( 1) ( 1)10
3
n n nnx n x n
− +⇒ + + =
2
2 310
3
nx nx
⎛ ⎞−⇒ + + =⎜ ⎟
⎝ ⎠
ewyksa dk fn;k x;k vUrj = 1
⇒ |α − β| = 1
⇒ D = 1
⇒ 2 24( 31) 1
3n n− − =
blfy,] n = 11
47. fuEu dFku
(p→q)→[(~p→q)→q] :
(1) p→~q ds lerqY; gS
(2) ,d gsRokHkkl (fallacy) gS
(3) ,d iqu#fDr (tautology) gS
(4) ~p→q ds lerqY; gS
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy p q p q→ (~ )p q→ (~ ) p q→ → q ( ) p q→ → → → [(~ ) ]p q q
T T T T T T
T F F T F T
F T T T T T
F F T F T T
( )iqu:fDr
48. oØ ( 2)( 3) 6y x x x− − = + ds ml fcanq ij] tgk¡ oØ
y-vk dks dkVrh gS] [khapk x;k vfHkyac fuEu esa ls fdlfcanq ls gksdj tkrk gS\
(1)1 1,
2 3
⎛ ⎞−⎜ ⎟⎝ ⎠
(2)1 1,
2 3
⎛ ⎞⎜ ⎟⎝ ⎠
(3)1 1,
2 2
⎛ ⎞− −⎜ ⎟⎝ ⎠
(4)1 1,
2 2
⎛ ⎞⎜ ⎟⎝ ⎠
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy ( 2)( 3) 6y x x x− − = +
y-vk ij] x = 0, y = 1
vc] vodyu djus ijA
( 2)( 3) (2 5) 1dy
x x y xdx
− − + − =
(6) 1( 5) 1
61
6
dy
dx
dy
dx
+ − =
= =
vc vfHkyEc dh izo.krk = –1
vfHkyEc dk lehdj.k y – 1 = –1(x – 0)
y + x – 1 = 0 ... (i)
js[kk (i)] 1 1,
2 2
⎛ ⎞⎜ ⎟⎝ ⎠
ls xqtjrh gS
49. fdUgha rhu èkukRed okLrfod la[;kvksa a, b rFkk c dsfy,
9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c) gS] rks %
(1) a, b rFkk c lekarj Js<h esa gSa
(2) a, b rFkk c xq.kksÙkj Js<h esa gSa
(3) b, c rFkk a xq.kksÙkj Js<h esa gSa
(4) b, c rFkk a lekarj Js<h esa gSa
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
JEE Main Paper I Code C (In Hindi)
13
gygygygygy 2 2 29(25 ) 25( 3 ) 15 (3 )+ + − = +a b c ac b a c
⇒ 2 2 2(15 ) (3 ) (5 ) 45 15 75 0a b c ab bc ac+ + − − − =
⇒ 2 2 2(15 3 ) (3 5 ) (15 5 ) 0− + − + − =a b b c a c
;g laHko gksxk tc
15 3 0− =a b rFkk 3 5 0− =b c rFkk 15 5 0− =a c
15 3 5= =a b c
1 5 3= =a b c
∴ b, c, a l-Js- esa gSaA
50. ;fn fcanq P(1, –2, 3) dk lery 2x + 3y – 4z + 22 = 0 esa
og izfrcac tks js[kk 1 4 5
x y z= = ds lekarj gS] Q gS] rks PQ
cjkcj gS %
(1) 42
(2) 6 5
(3) 3 5
(4) 2 42
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy PQ dk lehdj.k, 1 2 3
1 4 5
x y z− + −= =
ekuk M, ( 1, 4 2, 5 3)λ + λ − λ + gS
P
M
Q
pw¡fd ;g 2x + 3y – 4z + 22 = 0 ij fLFkr gS
λ = 1
Q ds fy,, λ = 2
nwjh 2 2 22 1 4 5 2 42PQ = + + =
51. ;fn 2 25(tan cos ) 2cos 2 9,x x x− = + rks cos 4x dk
eku gS %
(1)2
9(2)
7
9−
(3)3
5− (4)
1
3
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy 5 tan2x = 9 cos2x + 7
5 sec2x – 5 = 9 cos2x + 7
ekuk cos2x = t
59 12= +t
t
9t2 + 12t – 5 = 0
1
3=t pw¡fd
5
3≠ −t
2 1cos
3=x , cos 2x = 2cos2x – 1
= 1
3−
cos4x = 2 cos2 2x – 1
= 2
19
−
= 7
9−
52. ekuk a, b, c ∈ R A ;fn f(x) = ax2 + bx + c ,slk gSfd a + b + c = 3 gS rFkk lHkh x, y ∈ R ds fy, f(x +
y) = f(x) + f(y) + xy gS] rks 10
1
( )n
f n
=∑ cjkcj gS %
(1) 190 (2) 255
(3) 330 (4) 165
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy pw¡fd] ( ) ( ) ( )+ = + +f x y f x f y xy
fn;k gS] (1) 3=f
1= =x y j[kus ij ⇒ (2) 2 (1) 1 7= + =f f
blh izdkj ls] 1, 2= =x y ⇒ (3) (1) (2) 2 12= + + =f f f
vc] 10
1
( )=∑n
f n = (1) (2) (3) ... (10)+ + + +f f f f
= 3 + 7 + 12 + 18 + ... = S (ekuk)
vc] 3 7 12 18 ...n n
S t= + + + + +
iqu% 1
3 7 12 ...n n n
S t t−= + + + + +
gesa 3 4 5 ...nt n= + + + in izkIr gksrs gSa
= ( 5)
2
+n n
vFkkZr~] Sn
= 1=
∑n
n
n
t
= 215
2+∑ ∑n n
= ( 1)( 8)
6
+ +n n n
blfy,] S10
= 10 11 18
3306
× × =
JEE Main Paper I Code C (In Hindi)
14
53. ,d lery tks fcanq (1, –1, –1) ls gksdj tkrk gS rFkk
ftldk vfHkyac nksuksa js[kkvksa 1 2 4
1 2 3
x y z− + −= =−
rFkk
2 1 7
2 1 1
x y z− + += =− −
ij yac gS] dh fcanq (1, 3, –7) ls
nwjh gS %
(1)5
83
(2)10
74
(3)20
74
(4)10
83
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy ekuk lery ( 1) ( 1) ( 1) 0a x b y c z− + + + + = gS
;g nh x;h js[kkvksa ds yEcor gS
a – 2b + 3c = 0
2a – b – c = 0
gy djus ij, a : b : c = 5 : 7 : 3
∴ lery 5x + 7y + 3z + 5 = 0 gS
bl lery ls (1, 3, –7) dh nwjh = 10
83
54. ;fn S, 'b' dh mu fofHkUu ekuksa dk leqPp; gS ftuds
fy, fuEu jSf[kd lehdj.k fudk;
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
dk dksbZ gy ugha gS] rks S :
(1) ,d ifjfer leqPp; gS ftlesa nks ;k vfèkd vo;o gSa
(2) ,d gh vo;o okyk leqPp; gS
(3) ,d fjDr leqPp; gS
(4) ,d vifjfer leqPp; gS
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy
1 1 1
1 1 0
1
a
a b
=
⇒ –(1 – a)2 = 0
⇒ a = 1
a = 1 ds fy,
lehdj.k (1) rFkk (2) le:i gSa vFkkZr~] x + y + z = 1
dksbZ gy ugha gksus ds fy, x + by + z = 0. rc
b = 1
55. ;fn 2 3
4 1A
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
gS] rks adj (3A2 + 12A) cjkcj gS %
(1)51 84
63 72
⎡ ⎤⎢ ⎥⎣ ⎦
(2)72 63
84 51
−⎡ ⎤⎢ ⎥−⎣ ⎦
(3)72 84
63 51
−⎡ ⎤⎢ ⎥−⎣ ⎦
(4)51 63
84 72
⎡ ⎤⎢ ⎥⎣ ⎦
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy2 3
4 1
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
A
2 3
4 1
A I
− λ −− λ =
− − λ
= (2 – 2λ – λ + λ2 ) – 12
2( ) 3 10f λ = λ − λ −
, ( )A f λ∵ dks lrq"V djrk gSA
∴∴∴∴∴ A2 – 3A –10I
= 0
A2 – 3A = 10I
3A2 – 9A =
30I
3A2 + 12A =
30I
+ 21A
JEE Main Paper I Code C (In Hindi)
15
30 0 42 63
0 30 84 21
72 63
84 51
−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
−⎡ ⎤= ⎢ ⎥⎢ ⎥−⎣ ⎦
251 63
adj(3 12 )84 72
A A
⎡ ⎤+ = ⎢ ⎥
⎢ ⎥⎣ ⎦
56. ,d vfrijoy; fcanq ( 2, 3)P ls gksdj tkrk gS] rFkk
mldh ukfHk;k¡ ( 2, 0)± ij gSa] rks vfrijoy; ds fcanq P ij
[khaph xbZ Li'kZjs[kk ftl fcanq ls gksdj tkrh gS] og gS %
(1) ( 3, 2)
(2) ( 2, 3)− −
(3) (3 2, 2 3)
(4) (2 2, 3 3)
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy2 2
2 21
x y
a b− =
2 24a b+ =
rFkk 2 2
2 31
a b− =
2 2
2 31
4 b b− =
−⇒ 2
3b =2
1a∴ =2
21
3
yx∴ − =
∴ ( 2, 3)P ij Li'kZ js[kk dk lehdj.k 2 13
yx − = gS
Li"Vr% ;g (2 2, 3 3) ls xqtjrk gS
57. ekuk k ,d ,slk iw.kk±d gS fd f=kHkqt] ftlds 'kh"kZ(k, –3k), (5, k) rFk k (–k, 2) g S a ] dk k s = ki Qy28 oxZ bdkbZ gS] rks f=kHkqt ds yac&dsUnz ftl fcanq ijgS] og gS %
(1)3
1, 4
⎛ ⎞−⎜ ⎟⎝ ⎠(2)
12,
2
⎛ ⎞⎜ ⎟⎝ ⎠
(3)1
2, 2
⎛ ⎞−⎜ ⎟⎝ ⎠(4)
31,
4
⎛ ⎞⎜ ⎟⎝ ⎠
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy ks=kiQy =
3 11
5 1 282
2 1
k k
k
k
−=
−
5 4 0
5 2 0 56
2 1
− −+ − = ±
−
k k
k k
k
2 2( 7 10) 4 20 56− + + + = ±k k k k
25 13 10 56+ + = ±k k
25 13 46 0+ − =k k
25 13 66 0+ + =K K
25 13 46 0+ − =k k
k = 13 169 920
10
− ± +
= 13 1089
10
− ±
= 13 33
10
− ±
= 2, –4.6 vLohdk;Z
k = 2 ds fy,
A (2, –6)
C (–2, 2)D
m = 0
E
m= –2
B (5, 2)
m =
8
AD dk lehdj.k
x = 2 ...(i)
rFkk BE dk lehdj.k
12 ( 5)
2− = −y x
2 4 5y x− = −
2 1 0− − =x y ...(ii)
(i) o (ii) gy djus ij, 2y = 1
1
2=y
yEcdsUnz 1
2, 2
⎛ ⎞⎜ ⎟⎝ ⎠
gS
JEE Main Paper I Code C (In Hindi)
16
58. ,d iwQyksa dh D;kjh] tks ,d oÙk ds f=kT; [kaM ds :i esa gS]dh ?ksjkcanh djus ds fy, chl ehVj rkj miyCèk gSA rks iwQyksadh D;kjh dk vfèkdre ks=kiQy (oxZ eh- esa)] gS %
(1) 25
(2) 30
(3) 12.5
(4) 10
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy θr
r rθ
2 20r r+ θ = ... (i)
A = ks=kiQy = 2
2
2 2
rr
θ θ× π =π
... (ii)
220 2
2
r rA
r
−⎛ ⎞= ⎜ ⎟⎝ ⎠
2
220 210
2
r rA r r
⎛ ⎞−= = −⎜ ⎟⎝ ⎠
A ds vfèkdre gksus ds fy,
10 2 0 5dA
r rdr
= − = ⇒ =
2
22 0
d A
dr= − <
vr %r = 5 ds fy,] A vfèkdre gS
vc] 10 + θ·5 = 20 ⇒ θ = 2 (jsfM;u)
ks=kiQy = 22
5 25 sq m2
× π( ) =π
59. iQyu 1 1
: ,2 2
f R⎡ ⎤→ −⎢ ⎥⎣ ⎦
] tks
2( )
1
xf x
x=
+ kjk ifjHkkf"kr gS%
(1) vkPNknh gS ijUrq ,dSdh ugha gSA
(2) u rks vkPNknh vkSj u gh ,dSdh gSA
(3) O;qRØe.kh; gSA
(4) ,dSdh gS ijUrq vkPNknh ugha gSA
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy2
( )1
=+x
f xx
2 2
2 2 2 2
(1 ) 1 2 1( )
(1 ) (1 )
x x x xf x
x x
+ ⋅ − ⋅ −′ = =+ +
f′(x) dk fofHkUu vUrjkyksa esa fpg~u ifjorZu gksrk gSA
∴ ,dSdh ugha gS
21
xy
x=
+2
0yx x y− + =
y ≠ 0 ds fy,
2 1 11 4 0 , 0
2 2D y y
⎡ ⎤= − ≥ ⇒ ∈ − −⎢ ⎥⎣ ⎦
y = 0 ds fy, ⇒ x = 0
∴ ijkl dk Hkkx
∴ ijkl :1 1
, 2 2
⎡ ⎤−⎢ ⎥⎣ ⎦
∴ vkPNknh gS ijUrq ,dSdh ugha gSA
60. ,d O;fDr X ds 7 fe=k gSa] ftuesa 4 efgyk,¡ gSa rFkk 3 iq#"k
gSa] mldh iRuh Y ds Hkh 7 fe=k gSa] ftuesa 3 efgyk,¡ rFkk 4
iq#"k gSaA ;g ekuk x;k fd X rFkk Y dk dksbZ mHk;fu"B
(common) fe=k ugha gSaA rks mu rjhdksa dh la[;k ftuesa X
rFkk Y ,d lkFk 3 efgykvksa rFkk 3 iq#"kksa dks ikVhZ ij cqyk,a
fd X rFkk Y izR;sd ds rhu&rhu fe=k vk;sa] gS %
(1) 469 (2) 484
(3) 485 (4) 468
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy X(4 L 3 G) Y(3 L 4 G)
3 L 0 G 0 L 3 G
2 L 1 G 1 L 2 G
1 L 2 G 2 L 1 G
0 L 3 G 3 L 0 G
rjhdksa dh visfkr la[;k
= ( ) ( ) ( )2 2 24 4 4 3 4 3 3
3 3 2 1 1 2 3⋅ + ⋅ + ⋅ +C C C C C C C
= 16 + 324 + 144 + 1
= 485
JEE Main Paper I Code C (In Hindi)
17
PART–C : PHYSICS
61. ,d izskd izdk'k xfr dh vkèkh xfr ls 10 GHz vko`fÙk
ds ,d fLFkj lwe rjax (microwave) lzksr dh rjiQ tk
jgk gSA izskd kjk ekih x;h lwe rjax dh vko`fÙk dk
eku gksxk %
(izdk'k dh pky = 3 × 108 ms–1)
(1) 12.1 GHz (2) 17.3 GHz
(3) 15.3 GHz (4) 10.1 GHz
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy lkisfkdh; xfr ds fy,
f = 0
+−
c vf
c v; v = igq¡p dh lkisfkd pky
f = 210 10 3 17.3 GHz
2
+= =
−
cc
cc
62. fuEu izsk.kksa dks dsf'kdk fof/ ls ikuh dk i`"B ruko T
ukius ds fy;s fd;k tkrk gSA
dsf'kdk dk O;kl] D = 1.25 × 10–2 m
ikuh dk p<+ko] h = 1.45 × 10–2 m
g = 9.80 m/s2 rFk k ljyhd `r lEcU/ T =
310 N/m,
2
rhg × dks mi;ksx djrs gq, i`"B ruko esa
lEHkkfor =kqfV dk fudVre eku gksxk
(1) 1.5% (2) 2.4%
(3) 10% (4) 0.15%
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy 100 100 100Δ Δ Δ× = × + ×T D h
T D h
= 0.01 0.01
100 1001.25 1.45
× + ×
= 100 100
125 145+
= 0.8 + 0.689
= 1.489
1.5%
63. ,d v.kq ds dqN mQtkZ Lrjksa dks fp=k esa fn[kk;k x;k gSArjaxnSè;ks± ds vuqikr r = λ
1/λ
2 dk eku gksxk
–3E
–2E
– E
λ2
λ1
4
3E−
(1)2
3r = (2)
3
4r =
(3)1
3r = (4)
4
3r =
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy mQtkZ Lrj vkjs[k ls
1λ = hc
E
2
3
λ =⎛ ⎞⎜ ⎟⎝ ⎠
hc
E
∴ 1
2
1
3
λ=
λ
64. m = 10–2kg nzO;eku dk ,d fi.M ,d ekè;e esa tk jgk
gS vkSj ,d ?k"kZ.k cy F = –kv2 dk vuqHko djrk gSA fi.M
dk izkjfEHkd osx v0 = 10 ms–1 gSA ;fn 10s ds ckn mldh
ÅtkZ 1
8mv
02 gS rks k dk eku gksxk %
(1) 10–3 kg s–1 (2) 10–4 kg m–1
(3) 10–1 kg m–1 s–1 (4) 10–3 kg m–1
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy
2
0
2
0
1
18
1 4
2
f
i
mvk
kmv
= =
1
2
f
i
v
v
=
0
2f
v
v =
JEE Main Paper I Code C (In Hindi)
18
2 mdvkv
dt− =
0
0
0
2
2
0
v
t
v
dv kdt
mv
−=∫ ∫
0
0
2
0
1
v
v
kt
v m
−⎡ ⎤− =⎢ ⎥⎣ ⎦
0
0 0
1 2 kt
v v m− = −
0
0
1 kt
v m− = −
0 0
mk
v t=
210
10 10
−=
×
= 10–4 kg m–1
65. fLFkj nkc rFkk fLFkj vk;ru ij fof'k"V Å"ek;sa Øe'k%
Cp rFkk C
v gSaA ik;k tkrk gS fd
gkbMªkstu ds fy;s] Cp – C
v = a
ukbVªkstu ds fy;s] Cp – C
v = b
a vkSj b ds chp dk lgh lEcU/ gksxk %
(1) a = b (2) a = 14 b
(3) a = 28 b (4)1
14a b=
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy ekuk fu;r nkc ij eksyj Å"ek /kfjrk = Xp
rFkk fu;r vk;ru ij eksyj Å"ek /kfjrk = Xv
Xp
– Xv
= R
MCp
– MCv
= R
Cp
– Cv
= R
M
gkbMªkstu ds fy,; a = 2
R
N2 ds fy,_ b =
28
R
a
b = 14
a = 14b
66. ,d f=kT;k R rFkk yEckbZ l ds ,d leku csyu dk mlds
vfHkyEc fHkktd ds lkisk tM+Ro vk?kw.kZ I gSA tM+Ro vk?kw.kZ
ds fuEure eku ds fy;s vuqikr l/R D;k gksxk\
(1)3
2(2) 1
(3)3
2(4)
3
2
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy 2R
l
2 2
4 12= + mR m
I
2
2
4 3
⎡ ⎤= +⎢ ⎥
⎢ ⎥⎣ ⎦
mI R
2
4 3
⎡ ⎤= +⎢ ⎥π⎢ ⎥⎣ ⎦
m v
2
20
4 3
−⎡ ⎤= + =⎢ ⎥π⎣ ⎦
dI m v
d
2
2
3=
π
v
32
3
π= v
32 2
3
ππ = R
2
2
3
2=
R
3
2=
R
67. ,d jsfM;ks,fDVo ukfHkd&A ftldh v¼Z&vk;q T gS] dkk; ,d ukfHkd&B esa gksrk gSA le; t = 0 ij dksbZ HkhukfHkd&B ugha gSA ,d le; t ij ukfHkdksa B rFkk A dhla[;k dk vuqikr 0.3 gS rks t dk eku gksxk %
(1)log1.3
log2
t T= (2) t = T log (1.3)
(3)log(1.3)
Tt = (4)
log2
2 log1.3
Tt =
JEE Main Paper I Code C (In Hindi)
19
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy 0 0
0
0.3
−λ
−λ
−=
t
t
N N e
N e
⇒ eλt = 1.3
∴ λt = ln 1.3
ln 2ln 1.3
⎛ ⎞ =⎜ ⎟⎝ ⎠
tT
ln(1.3).
ln 2=t T
log(1.3)
log2=t T
68. fuEufyf[kr esa ls dkSulk dFku xyr gS\
(1) ,d larqfyr OghVLVksu lsrq esa] lsy ,oa xSYosuksehVj dksvkil esa cnyus ij 'kwU; foksi fcUnq izHkkfor gksrk gSA
(2) ,d /kjk fu;a=kd dks foHko foHkktd dh rjg mi;ksxdj ldrs gSaA
(3) fdjpkWiQ dk frh; fu;e mQtkZ ds lajk.k dks n'kkZrk gSA
(4) OghVLVksu lsrq dh lqxzkghrk lcls vf/d rc gksrh gStc pkjksa izfrjks/ksa dk ifjek.k rqY; gksrk gSA
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy ,d larqfyr OghVLVksu lsrq esa] 'kwU; foksi fcUnq vifjofrZrjgrk gS ;|fi lsy rFkk xsYosuksehVj dks vUr%ifjofrZr fd;kx;k gSA
69. ,d fo|qr ifjiFk esa ,d 2 μF èkkfjrk ds laèkkfj=k dks 1.0
kV foHkokUrj ds fcUnqvksa ds chp yxkuk gSA 1 μF èkkfjrk
ds cgqr lkjs laèkkfj=k tks fd 300 V foHkokUrj rd ogu
dj ldrs gSa] miyCèk gSaA
mijksDr ifjiFk dks izkIr djus ds fy, U;wure fdrus
laèkkfj=kksa dh vko';drk gksxh\
(1) 16 (2) 24
(3) 32 (4) 2
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy fuEu leatu vko';d gksxk %
Js.kh esa pkj bl izdkj dh 'kk[kkvksa ds lkFk lekUrj esa 1 μF
ds 8 laèkkfj=k
12
3
8
12
3
8
12
8
12
3
8
1000 V
250 V 250 V 250 V 250 V
8 Fμ 8 Fμ 8 Fμ 8 Fμ
1000 V
70. fn;s x;s ifjiFk esa tc èkkjk fLFkjkoLFkk esa igq¡p tkrh gS rks
èkkfjrk C ds laèkkfj=k ij vkos'k dk eku gksxk%
r
r1
r2
E
C
(1) ( )1
2
rCE
r r+ (2) ( )2
2
rCE
r r+
(3) ( )1
1
rCE
r r+ (4) CE
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy LFkkbZ voLFkk esa] la/kfj=k ls /kjk dk izokg 'kwU; gksxkA
i
E
r2
r1
r
C
i = E
r r2
+
VC = i r
2C =
Er C
r r
2
2+
VC = CE
r
r r
2
2+
71.2 V 2 V 2 V
1 Ω 1 Ω 1 Ω
2 V 2 V 2 V
Åij fn;s x;s ifjiFk esa izR;sd izfrjksèk esa èkkjk dk eku gksxk %
(1) 0.25 A (2) 0.5 A
(3) 0 A (4) 1 A
JEE Main Paper I Code C (In Hindi)
20
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy izR;sd ywi esa foHkokUrj 'kwU; gSA
∴ dksbZ èkkjk izokfgr ugha gksxhA
72. vk;ke ekWMqyu esa T;koØh; okgd vko`fÙk dks ωc ls rFkk
flXuy vkofÙk dks ωm ls n'kkZrs gSaA flXuy dh cS.M pkSM+kbZ
(Δωm
) dks bl rjg pqurs gSa fd Δωm
<< ωc. fuEu esa ls
dkSulh vkofÙk ekMqfyr rjax esa ugha gksxh\
(1) ωc
(2) ωm + ω
c
(3) ωc – ω
m(4) ω
m
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy ekWMqfyr rjax dh vkofÙk ijkl gS
ωc ± ω
m
∴ pw¡fd ωc >> ω
m
∴ ωm
dh miskk dh tkrh gSA
73. n-p-n Vªk¡ftLVj ls cuk;s gq, ,d mHk;fu"B mRltZd izoèkZdifjiFk esa fuosf'kr rFkk fuxZr foHkoksa ds chp dykarj dkeku gksxk %
(1) 90°
(2) 135°
(3) 180°
(4) 45°
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy n-p-n VªkaftLVj ds fy, mHk;fu"B&mRltZd vfHkfoU;kl esa]fuxZr rFkk fuos'kh oksYVrk ds eè; dykUrj 180° gSA
74. 100 gm nzO;eku okyk rk¡cs ds ,d xksys dk rkieku T
gSA mls ,d 170 gm ikuh ls Hkjs gq, 100 gm ds rk¡cs
ds dSyksjhehVj] tksfd dejs ds rkieku ij gS] esa Mky fn;k
tkrk gSA rRi'pkr~ fudk; dk rkieku 75°C ik;k tkrk
gSA T dk eku gksxk %
(fn;k gS % dejs dk rkieku = 30°C, rk¡cs dh fof'k"V Å"ek= 0.1 cal/gm°C)
(1) 885°C
(2) 1250°C
(3) 825°C
(4) 800°C
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy 100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45
10t – 750 = 450 + 7650
10t = 1200 + 7650
10t = 8850
t = 885°C
75. ;ax ds ,d ff>jh iz;ksx esa] f>fj;ksa ds chp dh nwjh 0.5 mm
,oa insZ dh f>jh ls nwjh 150 cm gSA ,d izdk'k iqat] ftlesa650 nm vkSj 520 nm dh nks rjaxnSè;Z gSa] dks insZ ijO;rhdj.k fÚUt cukus esa mi;ksx djrs gSaA mHk;fu"B dsUnzh;mfPp"B ls og fcUnq] tgk¡ nksuksa rjaxnSè;ks± dh nhIr fÚUtsalEikrh gksrh gSa] dh U;wure nwjh gksxh %
(1) 7.8 mm
(2) 9.75 mm
(3) 15.6 mm
(4) 1.56 mm
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy λ1 ds fy, λ
2 ds fy,
1λ
=m D
yd
2λ
=n D
yd
⇒2
1
4
5
λ= =
λm
n
λ1 ds fy,
1λ
=m D
yd
, λ1 = 650 nm
= 7.8 mm
76. ,d fo|qr fèkzqo dk fLFkj fèkzqo vk?kw.kZ p
gS tks fd
x-vk ls θ dks.k cukrk gSA fo|qr ks=k 1
ˆE Ei=
esa j[kus
ij ;g cy vk?kw.kZ 1
ˆT k= τ
dk vuqHko djrk gSA fo|qr
ks=k 2 1ˆ3E E j=
esa j[kus ij ;g cy vk?kw.kZ 2 1–T T=
dk vuqHko djrk gSA dks.k θ dk eku gksxk %
(1) 45°
(2) 60°
(3) 90°
(4) 30°
JEE Main Paper I Code C (In Hindi)
21
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy
pθ
y
x
z
cos sinp p i p j= θ + θ
1E Ei=
1 1T p E= ×
= ( )( cos sin )p i p j E iθ + θ ×
( )sink pE kτ = θ − ...(i)
2 13E E j=
2 1( cos sin ) 3T p i p j E j= θ + θ ×
13 cosk pE k−τ = θ ...(ii)
(i) o (ii) ls
sin 3 cospE pEθ = θ
tan 3θ =θ = 60°
77. ,d nzO;eku M ,oa yEckbZ l dh iryh ,oa ,d leku NM+
dk ,d fljk /qjkxzLr gS ftlls fd og ,d ÅèokZ/j lery
esa ?kwe ldrh gS (fp=k nsf[k;s)A /qjh dk ?k"kZ.k ux.; gSA
NM+ ds nwljs fljs dks /qjh ds Åij ÅèokZ/j j[kdj NksM+
fn;k tkrk gSA tc NM+ ÅèoZ ls θ dks.k cukrh gS rks mldk
dks.kh; Roj.k gksxk %
θ
z
x
(1)2
sin3
g θ
(2)3
cos2
g θ
(3)2
cos3
g θ
(4)3
sin2
g θ
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy θ dks.k ij cyk?kw.kZ
sin2
Mgτ = θ ⋅
θθMg
( , )M l
ω α ,
τ = Iα
Iα = sin2
Mg θ ∴
2
3
MI =
2
sin3 2
MMg⋅ α = θ
sin
3 2g
α θ=
3 sin
2
g θα =
78. 0°C ij j[ks gq, ,d ?ku ij ,d ncko P yxk;k tkrk gS
ftlls og lHkh rjiQ ls cjkcj laihfMr gksrk gSA ?ku ds
inkFkZ dk vk;ru izR;kLFkrk xq.kkad K ,oa js[kh; izlkj xq.kkad
α gSA ;fn ?ku dks xeZ djds ewy vkdkj esa ykuk gS rks
mlds rkieku dks fdruk c<+kuk iM+sxk\
(1)P
Kα (2)3
PK
α
(3) 3PKα (4)3
P
Kα
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy K = –
P
V
V
ΔΔ⎛ ⎞
⎜ ⎟⎝ ⎠
V
V
Δ=P
K
γ α = 3∴ V = V0 (1 + γΔt)
0
V
V
Δ = γΔt
∴ P
K = γΔt ⇒ Δt =
P
Kγ =
3
P
Kα
JEE Main Paper I Code C (In Hindi)
22
79. ,d 25 cm ifjek.k dh iQksdl nwjh ds vilkjh ysUl dks,d 20 cm ifjek.k dh iQksdl nwjh ds vfHklkjh ysUl ls15 cm dh nwjh ij j[kk tkrk gSA ,d lekarj izdk'k iqatvilkjh ysUl ij vkifrr gksrk gSA ifj.kkeh izfrfcEc gksxk %
(1) vkHkklh vkSj vfHklkjh ysUl ls 40 cm nwjh ij
(2) okLrfod vkSj vilkjh ysUl ls 40 cm nwjh ij
(3) okLrfod vkSj vfHklkjh ysUl ls 6 cm nwjh ij
(4) okLrfod vkSj vfHklkjh ysUl ls 40 cm nwjh ij
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygyf1 = 25 cm f
2= 20 cm
15 cm25 cm
I1
vfHklkjh ysUl ds fy,
u = –40 cm tks 2f ds cjkcj gS
∴ izfrfcEc okLrfod gksxk rFkk vfHklkjh ysUl ls 40 cm
dh nwjh ij gksxkA
80. X-fdj.ksa mRiUu djus ds fy;s ,d bySDVªkWu fdj.kiq¡t dks
foHkokUrj V ls Rofjr djds èkkrq dh IysV ij vkifrr fd;k
tkrk gSA blls fofoDr (characteristic) ,oa vfojr
(continuous) X-fdj.ksa mRiUu gksrh gSaA ;fn X-fdj.k LisDVªe
esa U;wure laHko rjaxnSè;Z λmin
gS rks log λmin
dk log V
ds lkFk cnyko fdl fp=k esa lgh fn[kk;k x;k gS\
(1)log λ
min
log V
(2)log λ
min
log V
(3)log λ
min
log V
(4)log λ
min
log V
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy X-fdj.k uyh esa]
min
hc
eVλ =
minln ln ln
hcV
e
⎛ ⎞λ = −⎜ ⎟⎝ ⎠
<ky Í.kkRed gS
y-vk ij vUr%[k.M /ukRed gSA
log V
log λmin
81. lw;Z dh fdj.kksa ls ,d [kqys gq, 30 m3 vk;ru okys dejs
dk rkieku 17°C ls c<+dj 27°C gks tkrk gSA dejs ds
vUnj ok;qeaMyh; nkc 1×105 Pa gh jgrk gSA ;fn dejs
ds vUnj v.kqvksa dh la[;k xeZ gksus ls igys ,oa ckn esa
Øe'k% ni o n
f gSa rks n
f – n
i dk eku gksxk %
(1) 1.38 × 1023 (2) 2.5 × 1025
(3) –2.5 × 1025 (4) –1.61 × 1023
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy n1
= eksyksa dh izkjfEHkd la[;k
n1 =
5
31 1
1
10 301.24 10
8.3 290
×= ≈ ××
PV
RT
JEE Main Paper I Code C (In Hindi)
23
n2
= eksyksa dh vfUre la[;k
= 5
32 2
2
10 301.20 10
8.3 300
×= ≈ ××
P V
RT
v.kqvksa dh la[;k dk ifjorZu
nf – n
i = (n
2 – n
1) × 6.023 × 1023
≈ – 2.5 × 1025
82. pqEcdh; ÝyDl ds cnyus ls 100 Ω izfrjksèk dh dq.Myh
esa izsfjr èkkjk dks fp=k esa n'kkZ;k x;k gSA dq.Myh ls xqtjus
okys ÝyDl esa cnyko dk ifjek.k gksxk %
(1) 225 Wb
(2) 250 Wb
(3) 275 Wb
(4) 200 Wb
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy ε = d
dt
φ
iR = d
dt
φ
d R idtφ =∫ ∫
ÝyDl esa ifjorZu dk ifjek.k = R × (/kjk rFkk le; xzkiQ
kjk ifjc¼ ks=kiQy)
= 100 × 1
2 ×
1
2 × 10
= 250 Wb
83. 15Ω ds dq.Myh izfrjksèk ds xSYosuksehVj ls tc 5 mA
dh èkkjk izokfgr dh tkrh gS rks og iw.kZ Ldsy foksi n'kkZrk
gSA bls 0–10V ijkl ds foHkoekih esa cnyus ds fy, fdl
eku ds izfrjksèk dks xSYosuksehVj ds lkFk Js.kh Øe esa
yxkuk gksxk\
(1) 2.045 × 103 Ω (2) 2.535 × 103 Ω
(3) 4.005 × 103 Ω (4) 1.985 × 103 Ω
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy ig = 5 × 10–3 A
G = 15 Ω
ekuk Js.khØe izfrjks/ R gSA
V = ig
(R + G)
10 = 5 × 10–3 (R + 15)
R = 2000 – 15 = 1985 = 1.985 × 103 Ω
84. 1 kg nzO;eku dk ,d d.k] ,d le; ij fuHkZj (time
dependent) cy F = 6t dk vuqHko djrk gSA ;fn d.k
fojkekoLFkk ls pyrk gS rks igys 1 s esa cy kjk fd;k x;k
dk;Z gksxk
(1) 22 J (2) 9 J
(3) 18 J (4) 4.5 J
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy 6 1dv
tdt
= ⋅
0
6
v
dv t dt=∫ ∫
12
0
62
tv
⎡ ⎤= ⎢ ⎥
⎣ ⎦
= 3 ms–1
W = ΔKE 1
1 9 4.5 J2
= × × =
85. ,d pqEcdh; vk?kw.kZ 6.7 × 10–2 Am2 ,oa tM+Ro vk?kw.kZ
7.5 × 10–6 kg m2 okyh pqEcdh; lqbZ ,d 0.01 T rhozrk
ds pqEcdh; ks=k esa ljy vkorZ nksyu dj jgh gSA 10 iwjs
nksyu dk le; gksxk
(1) 8.89 s
(2) 6.98 s
(3) 8.76 s
(4) 6.65 s
JEE Main Paper I Code C (In Hindi)
24
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
gygygygygy T = 2πI
MB
= 2π–6
–2
7.5 10
6.7 10 0.01
×× ×
= 2
1.0610
π ×
10 nksyuksa ds fy,]
t = 10T = 2π × 1.06
= 6.6568 ≈ 6.65 s
86. iFoh ds dsUnz ls nwjh d ds lkFk xq:Roh; Roj.k g dk cnyko
fuEu esa ls fdl xzkiQ esa lcls lgh n'kkZ;k x;k gS\
(R = iFoh dh f=kT;k)
(1)
O R
d
g
(2)
O R
d
g
(3)
O R
d
g
(4)
g
d
O
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy
O
gs
g
d = Rd
iFoh dh lrg ds vUnj g dk ifjorZu
2< = = ⋅Gm
d R g dR
2= = =
s
Gmd R g
R
2
Gmd R g
d> = =
87. ,d fi.M dks ÅèokZ/j Åij dh rjiQ iQsadk tkrk gSA fuEu
esa ls dkSu lk xzkiQ le; ds lkFk osx dks lgh n'kkZrk gS\
(1)
v
t
(2)
v
t
(3)
v
t
(4)
v
t
mÙkjmÙkjmÙkjmÙkjmÙkj (2)
gygygygygy Roj.k fu;r rFkk ½.kkRed gSA
v
t
JEE Main Paper I Code C (In Hindi)
25
88. nzO;eku m ,oa vkjfEHkd osx v ds ,d d.k dh VDdj
nzO;eku 2
m
ds fLFkj d.k&B ls gksrh gSA ;g VDdj lEeq[k
,oa izR;kLFk gSA VDdj ds ckn fM&czkWXyh rjaxnSè;ks± λA ,oa
λB dk vuqikr gksxk %
(1) 2A
B
λ=
λ (2)2
3
A
B
λ=
λ
(3)1
2
A
B
λ=
λ (4)1
3
A
B
λ=
λ
mÙkjmÙkjmÙkjmÙkjmÙkj (1)
gygygygygy 1 2
1
1 2
( )0
−= +
+m m v
v
m m
1
22
=
=
m m
mm
= 3
v
∴ 1.3
⎡ ⎤= ⎢ ⎥⎣ ⎦
vp m
1
2
1 2
20= +
+mv
v
m m
= 4
3
v
2
4 2
2 3 3
⎡ ⎤= =⎢ ⎥⎣ ⎦
m v mvp
∴ Mh&czksxyh rjaxnSè;Z2
1
2 :1A
B
p
p
λ= =
λ
89. ,d d.k] vkorZdky T ls ljy vkorZ xfr dj jgk gSA le;
t = 0 ij og lkE;koLFkk dh fLFkfr esa gSA fuEu esa ls dkSu lk
xzkiQ le; ds lkFk xfrt ÅtkZ dks lgh n'kkZrk gS\
(1)T0
(2)TT
2
0
(3)T
4
0 TT
2
(4)T TT
2
0
mÙkjmÙkjmÙkjmÙkjmÙkj (3)
gygygygygy 2 2 21K.E cos
2m A t= ω ω
T
4
T
2
90. ,d euq";] ,d fo'kkydk; ekuo esa bl izdkj ifjofrZr
gksrk gS fd mldh js[kh; foek;sa 9 xquk c<+ tkrh gSaA ekuk
fd mlds ?kuRo esa dksbZ ifjorZu ugha gksrk gS rks mlds
Vk¡x esa izfrcy fdrus xquk gks tk;sxk\
(1)1
9(2) 81
(3)1
81(4) 9
mÙkjmÙkjmÙkjmÙkjmÙkj (4)
Sol. 39
f
i
v
v
=
∵ ?kuRo leku jgrk gS
blfy, nzO;eku ∝ vk;ru
39
f
i
m
m
=
ks=kiQy
k=s kiQy
( )
( )
f
i
= 92
nOz ;ekuizfrcy
ks=kiQy( ) g×=
2
1
f i
i f
m A
m A
⎛ ⎞ ⎛ ⎞σ= ⎜ ⎟ ⎜ ⎟σ ⎝ ⎠ ⎝ ⎠
3
2
99
9= =
JEE Main Paper I Code C (In Hindi)