JEE MAIN EXAMINATION - 2014
QUESTIONS WITH SOLUTIONS
[HINDI VERSION]
PAPER CODE - H
Fastest Growing Institute of Kota (Raj.)
FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | CBSE | SAT | NTSE | OLYMPIADS
JEE MAIN Examination(2014) (Code - H) (Page # 1)
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[PHYSICS]3. yEckbZ dh , d vfor kU; Mksj h l s c¡/ks nzO; eku m ds , dckc dks , d Å/okZ/kj vk/kkj l s yVdk; k t kr k gSaA ckc Å/okZ/kj i jdks.kh; pky rad/s l s , d {kSfr t oÙk esa ?kw.kZu dj r k gSA fuyacufcUnq i j %(1) dks.kh; l aosx fn' kk esa i fj or Zu' khy gS i j Ur q i fj ek.k esa ugha(2) dks.kh; l aosx nksuksa fn' kk , oa i fj ek.k esa i fj or Zu' khy gSA(3) dks.kh; l aosx l aj f{kr j gr k gSA(4) dks.kh; l aosx i fj ek.k esa i fj or Zu' khy gSa i j Ur q fn' kk esa ughaASol. 1
L
L
L l fn' k fn' kk esa ?kwer k gS fdar q i fj ek.k fu; r j gr k gSA
4. , d Mk; kssM dh /kkj k&oksYVr k l EcU/k I = (e1000V/T – 1)mA l s nh t kr h gSa] t gk¡ V yxkbZ xbZ oksYVr k oksYV esa gS vkSj r ki ekuT fMxzh dSfYou esa gSA ; fn , d fo| kFkhZ 300 K i j 5 mA /kkj k uki r sgq; s eki us esa 0.01 V dh =kqfV dj r k gS] r c /kkj k ds eku esa mA esa=kqfV D; k gksxh\(1) 0.5 mA (2) 0.05 mA(3) 0.2 mA (4) 0.02 mASol. 3
I =
1e T
V1000
mA
TV1000
e = I + 1
log (I + 1) = T
V1000
1I1
× dI = 1000
T × dv
dI = 3001000
× (5 + 1) mA × 0.01
= 0.2 mA5. , d [ kqyh dk¡p dh uyh dks i kj s esa bl i zdkj Mqcks; k t kr kgSa fd i kj s ds Lr j l s 8 cm Åi j dk¡p dh uyh dh yEckbZ gSA uyhds [ kqys fl j s dks vc cUn dj l hy dj fn; k t kr k gS vkSj uyh dksÅ/okZ/kj vfr fj Dr 46 cm l s Åi j mBk; k t kr k gSA uyh esa i kj s dsÅi j ok; q Lr EHk dh yEckbZ vc D; k gksxh\(ok; qea.Myh; nkc = Hg dk 76 cm)(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm
1. 10 cm yEckbZ ds , d LVhy ds r kj ds fl j ksa i j t cr ki eku esa of) 100ºC dh t krh gSa r c bl dh yEckbZ fLFkj j [ kus dsfy, fl j ksa i j yxk; k x; k nkc gS %(LVhy dk ; ax i zR; kLFkr k xq.kkad 2 × 1011 N m–2 vkSj j sf[ kd i zl kjxq.kkad 1.1 × 10–5 K–1)(1) 2.2 × 107 i kLdy (2) 2.2 × 106 i kLdy(3) 2.2 × 108 i kLdy (4) 2.2 × 109 i kLdySol. 3
y = 2 × 1011
lf = l0 (1 + t) t = 100° C = 1.1 × 10–5 k–1
lf – l0 = t l = 1.1 × 10–5 × 100= 1.1 × 10–3
y = 3
p1.1 10
2 × 1011 = 3
p1.1 10
p = 2 × 1.1 × 1011 × 10–3
p = 2.2 × 108 Pa
2. , d l qpkyd z-v{k ds l kFk – 1.5 z < 1.5 m i j j [ kk
gS vkSj bl esa - za fn' kk esa fLFkj /kkj k 10.0 A i zokfgr gks j gh gSA ¼fp=k
ns[ ksa½A {ks=k B
= 3.0 × 10–4 e–0.2x ya T ds fy, ] l qpkyd dks
fLFkj pky l s x = 2.0 m, y = 0 m r c 5 × 10–3 s esa xfr dj kusds fy, vko' ; d ' kfDr dh x.kuk dh dhft , A x-v{k i j l ekUr j xfreku ysaA
(1) 14.85 W (2) 29.7 W (3) 1.57 W (4) 2.97 WSol. 4
w = dxe103310 x2.04
= 4. 5 × 10–2 (1 – e–0.4)
P = Tw
= 2.97
JEE MAIN Examination(2014) (Code - H)(Page # 2)
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Sol. 3
8cm
P0
76 × 8 = × pgas
76 8
= pgas
76 8
+ (54 – ) = 76
54 – = 7681
= 76 8
54 – 2 = 76 – 6082 + 22 – 608 = 0
= 222 (22) 4 808
2
= 2
5422 = 16 cm
6. l wph -I (fo| qr pqEcdh; r j ax i zdkj ) dks l wph -II (bul sl Ecf) r @vuqi z; ksfxr ) l s l qesfyr dhft , vkSj l wfp; ksa ds uhps fn; sx; s fodYi ksa esa l s l gh fodYi pqfu; s %
(a) voj Dr r j axs (i)ek¡l i sf' k; ksa dh fodfr
ds bykt ds fy,
(b) j sfM; ks r j axs (ii) i zl kj .k ds fy,
(c) , Dl -fdj .ksa (iii)gfì ; ksa ds vfLFkHkax
dh i gpku ds fy ,
(d) i j kcSaxuh fdj .ksa (iv)okrkoj .k dh vkst ksau
i j r } kj k vo' kks"k.k
l wph-I l wph-II
(a) (b) (c) (d)(1) (iii) (ii) (i) (iv)(2) (i) (ii) (iii) (iv)(3) (iv) (iii) (ii) (i)(4) (i) (ii) (iv) (iii)Sol. 2
7. nks oÙkh; IysVksa] ft uds chp nwjh 5 mm gSa] l s , d l ekUr j i fêdkl a/kkfj =k cuk; k x; k gS ft l ds chp i j koS| qr fLFkj kad 2.2 dk , di j koS| qr j [ kk x; k gS t c i j koS| qr esa fo| qr {ks=k3 × 104 V/m gS] r c /kukRed IysV dk vkos' k ?kuRo yxHkx gksxk%(1) 3 × 104 C/m2
(2) 6 × 104 C/m2
(3) 6 × 10–7 C/m2
(4) 3 × 10–7 C/m2
Sol. 3Q = cv
Q = d
A. r0 V
= AQ
= 0 r.d
V
= 0 r .E= 8.85 × 10–12 × 2.2 × 3 × 104
= 58.41 × 10–8 _~ 6 × 10–7 c/m2
8. , d fo| kFkhZ us , d NM+ dh yEckbZ eki dj 3.50 cm fy[ khAbl dks eki us esa ml us fdl mi dj .k dk i z; ksx fd; k\(1) , d LØw xst ft l ds ofuZ; j Ldsy esa 100 Hkkx gSa vkSj fi p 1mm gSA(2) , d Ldwy xst ft l ds ofuZ; j Ldsy esa 50 Hkkx gSa vkSj fi p1 mmgSA(3) , d ehVj LdsyA(4) , d ofuZ; j dSfyi l Z t gk¡ ofuZ; j Ldsy ds 10 Hkkx eq[ ; Ldsyds 9 Hkkxksa l s feyr s gSa vkSj eq[ ; Ldsy ds 1 cm esa 10 Hkkx gSaASol. 4
ofuZ; j dsfy i j dk vYi r ekad = 10
cm1.0
= 0.01 cmr Fkk 3.50 cm dk U; wur e i kB~; kad 0.01 cm gS
9. i zR; sd nzO; eku M ds pkj d.k t ks fd , d nwl j s l s l ekunwj h i j gSa] , d nwl j s ds vU; ksU; xq: Rokd"kZ.k i zHkko esa f=kT; k R ds , doÙk i j xfr ' khy gSA i zR; sd d.k dh pky gS %
(1) GM 1 2 2R
(2) 1 GM 1 2 22 R
(3) GMR
(4) GM2 2R
Sol. 2
JEE MAIN Examination(2014) (Code - H) (Page # 3)
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(2)(1)
(3) (4)
r
rmv2
= F14 + F24 + F34
= 2
2
r4Gm
+ 2
2
r2Gm
2
= 2
2
r2Gm
2
21
2mvr
= 2
2
r2Gm
2
221
v = Gm(1 2 2)4r
v = 21
)221(r
Gm
10. , d cgr Hkou esa] 40 W ds 15 cYc] 100 W ds 5 cYc]80 W ds 5 i a[ ksa , oa 1 kW dk 1 ghVj gSaA fct yh ds esUl dhoksYVr k 220 V gSA Hkou ds eq[ ; ¶; wt dh U; wur e {ker k gksxh %(1) 12 A (2) 14 A(3) 8 A (4) 10 ASol. 1
P = RV2
R1 = 40
)200( 2
dqy fo| qr /kkj k
= 1R
V × 15 +
2RV
× 5 + 3R
V × 5 +
4RV
× 1
R = V2/P
= 220
1540 +
2205100
+ 220
580 +
2201000
= 11.3626 AU; wur e eku 12 A.
11. , d d.k , d l j y j s[ kk esa l j y vkor Z xfr l s xfr ' khy gSA; g foj kekoLFkk l s i zkj EHk dj i zFke l Sfd.M esa nwj h a vkSj vxys l Sd.M esa nwj h 2a ml h fn' kk esa r ; dj r k gSA r c %(1) xfr dk vk; ke 4a gSA(2) nksyuksa dk vkor Z dky 6gSA(3) xfr dk vk; ke 3a gSA(4) nksyuksa dk vkor Z dky 8gSASol. 2
a
3a
A – A cos () = a ....(1)A – A cos (2) = 3a2A (1 – cos2 ) = 3a ...(2)l ehdj .k (2) dks (1) dk Hkkx nsus i j2 (1 + cos ) = 3cos = 1/2= /3
T2
= /3
T = 6
12. , d NksVs pqEcd dh fuxzkfgr k] t gk¡ yksgpqEcd vpqEcdh; gkst kr k gS] 3 × 103 A m–1 gSA pØksa dh l a[ ; k 100 , oa yEckbZ 10cm dh , d i fj ukfydk l s i zokfgr vko' ; d /kkj k dk eku] ft l l s fdpqEcd i fj ukfydk ds vUnj gksus i j vpqEcdh; gks t k; s] gS %(1) 3 A (2) 6 A(3) 30 mA (4) 60 mASol. 1
fuxzkfgr k 0
B = 3 × 103 A/m
B = 0 (3 × 103) = 0 nI
= 0
1.0
100 I
I = 100
1.0103 3 = 3 A
JEE MAIN Examination(2014) (Code - H)(Page # 4)
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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13. vxzfl r ck; l okyk Mk; ksM+ t ksM+ gS %
(1)
(2)
(3)
(4)
Sol. 3
+2V
( )mPp ( )fuEu
–2V
vxz vfHkufr
14. , d ek/; e esa fo| qr pqEcdh; r j axksa ds l apj .k ds nkSj ku %(1) fo| qr h; Åt kZ ?kuRo pqEcdh; Åt kZ ?kuRo ds cj kcj gSA(2) nksuksa fo| qr h; , oa pqEcdh; Åt kZ ?kuRo ' kwU; gSA(3) fo| qr h; Åt kZ ?kuRo pqEcdh; Åt kZ ?kuRo dk nksxquk gSA(4) fo| qr h; Åt kZ ?kuRo pqEcdh; Åt kZ ?kuRo dk vk/kk gSASol. 1
, d fo| qr pqEcdh; r j ax esa
1/2 0 E2 =
0
2
2B
nksuksa l eku gksr s gS
15. ; gk¡ n' kkZ; s x; s i fj i Fk esa] fcUnq 'C' dks fcUnq 'A' l s r c r dt ksM+s j [ kk t kr k gS t c rd fd i fj i Fk esa i zokfgr /kkj k fLFkj u gks t k, Ar Ri ' pkr ~] vpkud] fcUnq 'C' dks fcUnq 'A' l s gVkdj fcUnq 'B' l s t =0 l e; i j t ksM+ fn; k t kr k gSA t = L/R i j i zfr j ks/k i j oksYVr k dki zsj dRo i j oksYVr k l s vuqi kr gksxk %
(1) –1 (2) 1 e
e
(3) e
1 e(4) 1
Sol. 1t c BC dks t ksM+k t kr k gS] r c L , oa R i fj i Fk es j gt kr s gS
R
– +
+
–L
KVL yxkus i j , VR + VL = 0 VR = –VL
L
R
VV
= – 1
16. f=kT; k R , oa nzO; eku m ds , d , dl eku [ kks[ kys csyu dspkj ksa r j Q , d nzO; ekufoghu Mksj h l s , d nzO; eku 'm' voyafcr gSaA; fn Mksj h csyu i j fQl yr h ugha gSa] r c NksM+s t kus i j nzO; eku fdlRoj .k l s fxj sxk\
(1) 6g5
(2) g (3) 3g2
(4) 2g
Sol. 4mg – T = ma .....(1)TR = I
TR = mR2 aR
T = ma ......(2) T
m
mg – ma = maa = g/2
17. f} i j ek.kqd vkn' kZ xSl dk , d eksy pØh; i zfØ; k ABC l sxqt j r k gS t Sl k fd fp=k esa n' kkZ; k x; k gSA i zfØ; k BC : ) ks"e gSA A,B , oa C ds r ki eku Øe' k% 400K, 800K , oa 600K gSaA l gh dFkupqfu; s %
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(1) i zfØ; k AB esa vkUr fj d Åt kZ esa i fj or Zu -350 R gSA(2) i zfØ; k BC esa vkUr fj d Åt kZ esa i fj or Zu -500 R gSA(3) l Ei w.kZ pØh; i zfØ; k esa vkUr fj d Åt kZ esa i fj or Zu 250 R gSA(4) i zfØ; k CA esa vkUr fj d Åt kZ esa i fj or Zu 700 R gSASol. 2
C600
800B
A400
P
V
(1) dUAB = n cv dT
= 1 × 2R5
(400)
= 1000 R(2) dUBC = n cv dT
= 1 × 2R5
(–200)
= – 500 R(3) dUABCA = 0
(4) dUCA = 1 × 2R5
(–200) = – 500 R
18. Å¡pkbZ H dh , d ehukj l s] pky u l s , d d.k dksÅ/okZ/kj Åi j dh vksj Qsadk t kr k gSA d.k dks i Foh rd fxj us esa yxkl e; ml ds mPpr e fcUnq r d i gq¡pus ds l e; dk n xquk gSaA
H, u , oa n ds chp l EcU/k gSA(1) 2g H = nu2(n-2) (2) g H = (n-2)u2
(3) 2g H = n2u2 (4) g H = (n-2)2u2
Sol. 1
u
H
–H = ut – 1/2 g t2
t' = u/g
–H = u
gnu
– 1/2 g 2
gnu
–H = gnu2
– 1/2 gun 22
–H = gnu2
(1 – 21
n)
–2Hg = nu2 (2 – n) 2Hg = nu2 (n – 2)
19. Økmu dk¡p
23
l s cus , d i r ys mÙky ysUl dh
Qksdl yEckbZ f gSA t c bl s vi or Zukad 34
, oa 35
okys nks fHkUu nzoksa
esa j [ kdj eki k t kr k gS] r c Qksdl yEckb; k¡ Øe' k f1 , oa f2 gSaAQksdl yEckb; ksa ds chp l gh l EcU/k gSa %(1) f2 > f vkSj f1 _ .kkRed gks t kr k gSA(2) f1 , oa f2 nksuksa _ .kkRed gks t kr s gSA(3) f1 = f2 < f(4) f1 > f vkSj f2 _ .kkRed gks t kr k gSASol. 4
pwafd 4/3 < 3/2 bl fy, f1 > fpwafd 5/3 > 3/2 bl fy, f2 dk fpaâ cnyr k gS
20. r k¡cs] i hr y , oa LVhy dh rhu NM+ksa dks Y- vkdkj l aj puk esa osYMfd; k x; k gSaA i zR; sd NM+ dh vuqi zLFk dkV dk {ks=kQy = 4cm2 gSAr k¡cs dh NM+ ds fl j s dk r ki eku 100°C gSa t cfd i hr y , oa LVhyds fl j s 0°C r ki eku i j j [ ks x; s gSaA r k¡cs] i hr y , oa LVhy dh NM+ksadh yEckbZ; k¡ Øe' k% 46, 13 , oa 12 cms gSaA NM+ksa dks] muds fl j ksadks NksM+dj ] okr koj .k l s Å"eh; j ks/kh fd; k x; k gSA r k¡cs] i hr y , oaLVhy dh Å"ek pkydr k, ¡ Øe' k% 0.92, 0.26 , oa 0.12 CGSbdkbZ gSaA r k¡cs dh NM+ l s i zokfgr Å"ek dh nj gS %(1) 4.8 cal/s (2) 6.0 cal/s(3) 1.2 cal/s (4) 2.4 cal/sSol. 1
100° 0°C
Cu46cmk=0.92
Br13 cmk=0.26T
st, 12 cmk = 0.12
0°C
JEE MAIN Examination(2014) (Code - H)(Page # 6)
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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fdj pksQ ds fu; e l s
1R100T
+ 2
T 0R
+ 3
T 0R
= 0
)46(100T
(0.92) + 13T
(0.26) + 12T
(0.12) = 0
50100T
+ 50T
+ 100T
= 0
2T – 200 + 2T + T = 0T = 40
Cu
dQdt
= 46
)40100( (0.92)(4) = 4.8 cal/s
21. yEckbZ 85 cm ds , d i kbi ds , d fl j s dks cUn dj fn; k t kr kgSa i kbi esa ok; q Lr EHk ds l EHko] i zkd fr d nksyuksa dh og l a[ ; kfudkfy; s ft udh vkofÙk 1250 Hz l s de gSaA ok; q esa /ofu dk osx340 m/s gSA(1) 6 (2) 4 (3) 12 (4) 8Sol. 1
f = L4V
= 210854340
= 100 Hz
100 H, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz = 6
22. , d oÙkkdkj uyh Å/okZ/kj r y esa gSA nks nzo] t ks , d nwl j s l sfefJr ugha gksr s r Fkk ft udk ?kuRo d1 , oa d2 gSa] uyh esa Hkj s x; s gSaAi zR; sd nzo dsUnz i j 90° dk dks.k var fj r dj r k gSaA muds var % i "B
dks t ksM+us okyh f=kT; k Å/okZ/kj l s dks.k cukr h gSA vuqi kr 1
2
dd gSa%
(1) 1 tan1 tan
(2) 1 sin1 cos
(3) 1 sin1 sin
(4) 1 cos1 cos
Sol. 1
R
Rsin
Rcos
R–RcosR-Rsin
RR
R – R sin = d1 (R – R sin ) + d2 (R sin + R cos )d1 (1 – sin ) = d1 (1 – cos ) + d2 (sin + cos )d1 – d1 sin = d1 – d1 cos + d2 (sin + cod )d1 (cos – sin ) = d2 (sin + cos )
2
1
dd
=
sincoscossin
=
tan1tan1
23. , d gj s j ax dk i zdk' k i kuh l s ok; q&t y vUr j ki "B i j ØkfUr ddks.k () l s vki fr r gSA l gh dFku pqfu; sA(1) n' ; i zdk' k dk og Li sDVªe] ft l dh r j axnS/; Z gj s i zdk' k l svf/kd gS] i kuh l s ok; q ds ek/; esa ckgj fudysxkA(2) n' ; i zdkj dk l Ei w.kZ Li sDVªe i kuh l s vfHkyEc l s fofHkUu dks.kksai j ckgj fudysxkA(3) vfHkyEc l s 90° dks.k i j i kuh l s n' ; i zdk' k dk l Ei w.kZ Li sDVªeckgj fudysxkA(4) n' ; i zdk' k dk og Li sDVªe] ft l dh r j axnS/; Z gj s i zdk' k l s degS] i kuh l s ok; q ds ek/; e esa ckgj fudysxkASol. 4
gj k
1
f > fgj k
< gjk
> gj k
T.I.R
24. gkbMªkst u (1H1), M~; wVsfj ; e (1H
2), , d/kk vk; fur ghfy; e(2He4)+ vkSj f} /kk vk; fud yhfFk; e (3Li6)++ l Hkh esa , d bysDVªkWuukfHkd ds pkj ksa vksj gSaA n = 2 l s n = 1 ds bysDVªkWu l aØe.k i jfopkj dhft , A ; fn mRl ft Zr fofdj .k dh r j axnS/; Z Øe' k% 1, 2,3
, oa 4 gSa] r c fuEufyf[ kr l EcU/kksa esa l s dkSu l k yxHkx l gh gS\(1) 1 = 2 = 43 = 94(2) 1 = 22 = 33 = 44(3) 41 = 22 = 23 = 4(4) 1 = 22 = 23 = 4
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Sol. 1
hc
= Rc z2
2
22
1 n1
n1
1
z2
1
1 (1)2,
2
1 (1)2,
3
1 (2)2,
4
1 (3)2
1 : 2 : 3 : 4 = 1 : 1 : 41
: 91
25. gkbMªkst u i j ek.kq ds 3 2 l aØe.k ds l axr fofdj .k , d /kkr qi "B i j vki fr r gksdj QksVksbysDVªkWu mRi Uu dj r k gSA ; s bysDVªkWu3×10-4T ds , d pqEcdh; {ks=k esa i zos' k dj r s gSA ; fn bysDVªkWuksa } kj kvuqxkeh vf/kdr e oÙkh; i Fk dh f=kT; k 10.0mm gks] r c /kkr q dkdk; Z Qyu yxHkx gS %(1) 0.8 ev (2) 1.6 ev(3) 1.8 ev (4) 1.1 evSol. 4
r = qBmv
v = m
qBr =
19 4 3
31
1.6 10 3 10 10 109.1 10
= 5.27 × 105
KEmax = 1/2 mv2
= 1/2 × 9.1 × 10–31 × (5.27 × 105)2
= 19
21
106.11036.126
= 78.975 × 10–2
= 0.78 eV
E = 13.6
2
22
1 n1
n1
= 13.6
91
41
= 1.88 ev1.88 = 0.78 + = 1.1 eV
26. , d i "B i j , d nzO; eku m dk CykWd j [ kk gSA i "B dh
Å/okZ/kj vuqi zLFk dkV y = 3x
6 l s nh t krh gSA ; fn ?k"kZ.k xq.kkad 0.5
gS] r c /kj r h l s Åi j og vf/kdre Å¡pkbZ] ft l i j fcuk fQl ys CykWdj [ kk t k l dr k gS] gSa %
(1) 13 m (2)
12
m
(3) 16 m (4)
23 m
Sol. 3
m
f
y = 6x3
dxdy
= 6x3 2
= 2x2
f = mg cos = mg sin = tan
0.5 = 2x2
x = 1
y = 6x3
= 61
m
27. t c , d j cM+ ds NYys dks x nwj h r d r kfur fd; k t kr k gS( r ci fj ek.k F = ax + bx2 , k , d i zR; u; u cy yxr k gS t gk¡ a , oa bfLFkj kad gSA fcuk r kfur j cM+ ds NYys dks L l s r kfur djus esa fd; kx; k dk; Z gS %
(1) 2 3aL bL
2 3 (2)
2 31 aL bL2 2 3
(3) aL2 + bL3 (4) 12
(aL2 + bL3)
JEE MAIN Examination(2014) (Code - H)(Page # 8)
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Sol. 1
W = Fdx
W = Fdx(ax + bx2) dx
W =
L
0
32
3bx
2ax
W =
3bL
2aL 32
28. i kuh dks xeZ dj us i j ] cr Zu dh r yh esa cqycqys cur s gSa vkSjfoyXu gksdj Åi j dh vksj mBr s gSaA cqycqyksa dks f=kT; k R dk xksykeku ysa vkSj cr Zu dh r yh esa oÙkh; Li ' kZ dh f=kT; k r ysaA ; fn r<<RvkSj i kuh dk i "B r uko T gSa] r c cqycyksa ds cl foyXu gksus l s t j ki gys r dk eku gSa %¼i kuh dk ?kuRo w gS½
(1) 2 WgR
T
(2) 2 W3 gR
T
(3) 2 WgR
3T
(4) 2 WgR
6T
Sol. 3
R T
r
34
R3 w g = P (r2) + T (2r) sin
34
R2 w g = RT2
(r2) + T (2r) Rr
34
R3 w g = RT4
r2
r = R2 T3gw
29. /kzqo.k ds vU; ks; yEcor ~ r yksa okys l er y /kzqoh; i zdk' k dh nksi qat A , oa B , d i ksyj k; M+ } kj k t krh gSA ml fLFkfr l s t gk¡ i qat Adh vf/kdr e r hozr k gS ¼vkSj i qat B dh ' kwU; r hozr k gS½ i ksyj k; M dk30° l s ?kw.kZu nksuksa i qat ksa dks , dl eku | qfr eku i zr hr gksr k gSA ; fn
nksuksa i q¡t ksa dh i zkj fEHkd r hozr k, ¡ Øe' k% IA , oa IB gSa] r c A
B
II dk eku
gS %
(1) 1 (2) 13
(3) 3 (4) 32
Sol. 2IA cos2 30° = IB cos2 60°
B
A
II
=
30cos60cos
2
2
= 2
2
)2/3()2/1(
= 4/34/1
= 1/3
30. eku ysa O; kse esa , d fo| qr {ks=k 2E 30x i gSA r c foHkokUr j
VA – VO t gk¡ VO ewyfcUnq i j foHko , oa VA, x = 2m i j foHko gSa]gS %(1) –80 J (2) 80 J(3) 120 J (4) –120 JSol. 1
dv = dr.E
VA – V0 = 2
0
2dxx30
= –
2
0
3
3x30
= – [10x3]= – 10 [8 – 0]VA – V0 = –80 J
JEE MAIN Examination(2014) (Code - H) (Page # 15)
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[MATHEMATICS]
SECTION – ASingle Correct
31. l er y 2x – y + z + 3 = 0 esa j s[ kk
31x
= 1
3y =
54z
ds i zfr fcac okyh j s[ kk gS%
(A) 3
3x =
15y
= 52z
(B) 33x
= 15y
= 5
2z
(C) 3
3x =
15y
= 52z
(D) 33x
= 15y
= 5
2z
Sol. [1]x 1 y 3 z 4
3 1 5
2 3 4 3x 1 y 3 z 4 22 1 1 4 1 1
x 1 y 3 z 4 22 1 1
x= 1 – 4, y = 3 + 2, z = 4 – 2P' (–3, 5 ,2)j s[ kk l er y ds l ekUr j gSai zfr fcEc j s[ kk dh l ehdj .k
x 3 y 5 z 23 1 5
32. ; fn (1 + ax + bx2) (1 – 2x)18 ds x dh ?kkrksa esa i zl kjesa x3 r Fkk x4 nksuksa dsd xq.kkad ' kwU; gS] r ks (a, b) cj kcjgS%
(A)
3
251,16 (B)
3
251,14
(C)
3
272,14 (D)
3
272,16
Sol. [4](1 + ax + bx2) (1 – 2x)18
x3 dk xq.kkad = 18C3 (–2)3 + a. 18C2 (–2)2
+ b . 18C1(–2)1
= 6528 a612 36b 0
x4 dk xq.kkad = 18C4 (–2)4 + a . 18C3 (–2)3 + b 18C2 (–2)2
= 48960 a.( 6528) b612 0
(16, 2723
) i zkIr dj r s gSaA
33. ; fn a R r Fkk l ehdj .k– 3(x – [x])2 + 2(x – [x]) + a2 = 0(t gk¡ [x] ml cM+s l s cM+s i w.kkZad dks n' kkZr k gS t ks x gS)dk dksbZ i w.kkZadh; gy ugh gS] r ks a ds l Hkh l aHko eku ft lvar j ky esa fLFkr gS] og gS%(A) (–1, 0) (0, 1) (B) (1, 2)(C) (–2, –1)(D) (–, –2) (2, )
Sol. [1]x – [x] = {x} = t [0, 1)–3t2+ 2t + a2 = 0 a2= 3t2 – 2t [0, 1)pwafd l ehdj .k i w.kkZadh; ewy ugha j [ kr h gSa : t 0 a2 (0, 1) a (–1, 0) (0, 1)
34. ; fn accbba
= 2cba gS] r ks cj kcj
gS%(A) 2 (B) 3(C) 0 (D) 1
Sol. [4]Li "V gSa= 1
35. i gyh 50 l e i zkd r l a[ ; kvksa dk i zl j .k gS%
(A) 4
833(B) 833
(C) 437 (D) 4
437
Sol. [2]
50 51 101V 4
6 50
50 51 50 514
50 50 4
= 17 [202 – 153] = 17 × 49 = 833
JEE MAIN Examination(2014) (Code - H)(Page # 16)
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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36. , d i {kh 20 eh- Å¡ps , d Å/okZ/kj [ kaHks ds f' k[ kj i j cSBkgS r Fkk bl dk Hkwfe ds , d fcanq O l s mUu; u dks.k 45º gSA; g i {kh O l s i j s {kSfr t fn' kk esa mM+r k gSA , d l sdaM dsckn] O l s i {kh dk mUu; u dks.k ?kV dj 30º j g t kr k gSAr ks (eh- i zfr l s- esa) i {kh dh pky gS :
(A) 40( 2 – 1) (B) 40( 3 – 2 )
(C) 20 2 (D) 20( 3 – 1)
Sol. [4]
tan30° = 20 1
x 20 3
x + 20 = 20 3
x = 20( 3 1)
37. l ekdy
0
2 dx2xsin4
2xsin41 cj kcj gS &
(A) – 4 (B) 32
– 4 – 4 3
(C) 4 3 – 4 (D) 4 3 – 4 – 3
Sol. [4]
0
x1 2sin dx2
= 3
0
x1 2sin dx2
3
x1 2sin dx2
= 3
03
x xx 4cos x 4cos2 2
= 4cos 0 43 6 4cos 4cos
2 3 6
= 4 3 43
38. dFku ~ (P ~q) gS :(A) p q ds r qY;(B) ~p q ds r qY;(C) , d i qu: fDr (tautology)(D) , d gsRokHkkl (fallacy)
Sol. [1]
p q ~ q p ~ q ~ p ~ qT T F F TT F T T FF T F T FF F T F T
p qTFFT
39. ; fn A , d , sl k 3 × 3 O; wRØe.kh; gS fd AA' = A'A r FkkB = A–1 A' gS, r ks BB' cj kcj gS%(A) 1 + B (B) 1(C) B – 1 (D) (B–1)'
Sol. [2]AA' = A'AB = A–1A'B' = A(A–1)'B . B' = A–1 (A' A) (A–1)'= A–1 (A A') (A–1)1
= (A–1 A) (A') (A')–1
= I
40. l ekdy dxex1x1 x
1x
cj kcj gS :
(A) (x – 1) ce x1x
(B) x ce x
1x
(C) (x + 1) ce x1x
(D) –x ce x
1x
Sol. [2]1 1x xx x1e x e dx
x
...(1)
1xxe
= f(x)
1xxe
2
11x
dx = f'(x)
1 1x+ x+x x
2f(x)
xf '(x)
1e x.e 1 dxx
x f(x) = x . 1x+xe + C
JEE MAIN Examination(2014) (Code - H) (Page # 17)
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41. ; fn z , d , sl h l fEEJ l a[ ; k gS fd |z| 2 gS, r ks
21z dk U; wur e eku :
(A) 25
ds cj kcj gSA
(B) var j ky (1, 2) esa fLFkr gSA
(C) 25
l s fuj ar j cM+k gSA
(D) 23
l s fuj ar j cM+k gS i j Ur q 25
l s de gSA
Sol. [2]|z| 2
1 1 1| z | z | z |2 2 2
min
1 3Z2 2
42. ; fn g Qyu f dk O; wRØe gS r Fkk f'(x) = 5x11
gS, r ks
g'(x) cj kcj gS%(A) 1 + x5 (B) 5x4
(C) 5)}x(g{11
(D) 1 + {g(x)}5
Sol. [4]f(g(x)) = xf'(g(x)) g'(x) = 1
g'(x) = 1
f '(g(x))
g'(x) = 5
11
1 (g(x))
1 + (g(x))5
43. ; fn 0, f(n) = n + n r Fkk
)4(f1)3(f1)2(f1)3(f1)2(f1)1(f1)2(f1)1(f13
= K(1– )2 (1 – )2 ( – )2 gS, r ks K cj kcj gS%
(A) (B) 1
(C) 1 (D) –1
Sol. [1]f(n) = n + n
3
1 1 1 1 11 1 1
1 1 1
= 2 2
1 1 1 1 1 11 11 1
= (1 – )2 ( – )2 ( – 1)2
k 1
44. ekuk fk(x) = k1
(sink x + cosk x) gS] t gk¡ x R
r Fkk k 1 gS] r ks f4(x) – f6(x) cj kcj gS%
(A) 61
(B) 31
(C) 41
(D) 121
Sol. [4]
f4(x) – f6(x) = 14
(sin4x + cos4x) – 16
(sin6 x + cos6 x)
= 14
(1 – 2sin2x cos2x) – 16 (1 – 3 sin2x cos2x)
= 1 1 3 2 14 6 12 12
45. ekuk r Fkk l ehdj .k px2 + qx + r = 0, p 0 ds
ewy gSA ; fn p, q, r l ekar j Js<+h esa gS r Fkk 1
+ 1
=
4 gS] r ks | – | dk eku gS%
(A) 961
(B) 9172
(C) 934
(D) 9132
JEE MAIN Examination(2014) (Code - H)(Page # 18)
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Sol. [4]
fn; k x; k gSa fd = –qp =
rp
p, q, r l ekUr j Js<h , r c
q – p = r – q 2q r p ...(1)
i qu% 1 1 4
= 4
q 4r ...(2)
l ehdj .k (1) o (2) l s– 8r = r + p
p 9r ...(3)
vc, | – | = 2( ) 4
= 2q 4pr|p |
= 2 216r 36r
|9r |
= 52 2 139 9
46. ekuk A r Fkk B nks , sl h ?kVuk, ¡ gSa fd )BA(P = 61
,
P (A B) = 41
r Fkk P( A ) = 41
gS t cfd A ?kVuk
A ds i wj d dks n' kkZr k gSA r ks ?kVuk, ¡ A r Fkk B :(A) i j Li j vi ot hZ r Fkk Lor a=k gSA(B) l el EHkkoh gSa i j Ur q Lor a=k ugha gSA(C) Lor a=k gS i j Ur q l el EHkkoh ugha gSA(D) Lor a=k gSa r Fkk l el EHkkoh gSaA
Sol. [3]
1P(A B)6
1P(A B)4
1P(A)4
Li "V gSa fd P(A)= 34
r Fkk P(A B) = 56
P(A) + P(B) – P(A B) = 56
34
+ P(B) – 1 54 6
P(B) = 5 3 16 4 4
= 13
P(A B) = 14
P(A) . P(B) = 3 1 14 3 4
bl fy, Lor a=k
P(A) P(B) bl fy, l Ei zkf; d ?kVuk, ¡ ugha gSa
47. ; fn f r Fkk g, [0, 1] esa vodyuh; Qyu gSa t ksf(0) = 2 = g(1), g(0) = 0 vkSj f(1) = 6 dks l ar q"Vdj r s gS] r ks fdl h c ]0, 1[ ds fy, %(A) 2f'(c) = g'(c)(B) 2f'(c) = 3g'(c)(C) f'(c) = g'(c)(D) f'(c) = 2g'(c)
Sol. [4]2g'(c) = f'(c)
= 2 g(1) g(0) f(1) f(0)
1 0 1 0
= 22 0 6 2
1 1
4 = 4
48. ekuk fdl h l e; t i j t hfor [ kj xks' kksa dh t ul a[ ; k vody
l ehdj .k dt
)t(dp =
21
p(t) – 200 } kj k fu; af=kr gSA
; fn p(0) = 100, gS] r ks p(t) cj kcj gS%(A) 400 – 300 et/2 (B) 300 – 200 e–t/2
(C) 600 – 500 et/2 (D) 400 – 300 e–t/2
JEE MAIN Examination(2014) (Code - H) (Page # 19)
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Sol. [1]
2dp(t) dtp(t) 400
2 log|p(t) – 400| = t + c ...(1)t = 0, p = 100
2 log (300) = C
l ehdj .k (1) l s2log|p(t) – 400| = t + 2 log (300)|p(t) – 400| = et/2 . elog(300)
t /2p(t) 400 e (300)
49. ekuk C , d oÙk gS ft l dk dsanz (1, 1) i j gS r Fkk f=kT; k= 1 gSA ; fn T dsUnz (0, y) okyk oÙk gS t ks ewy fcaanq l sgksdj t kr k gS r Fkk oÙk C dks cká : i l s Li ' kZ dj r k gS] r ksT dh f=kT; k cj kcj gS%
(A) 23
(B) 23
(C) 21
(D) 41
Sol. [4]
C1C = 1 + y(1 – 0)2 + (1 – y)2 = (1 + y)2
1 + 1 – 2y + y = 1 + y2 + 2y4y = 1
1y4
50. A = {(x, y) : x2 + y2 1 r Fkk y2 1 – x} ds } kj ki znÙk {ks=k dk {ks=kQy gS%
(A) 2
+ 34
(B) 2
– 34
(C) 2
– 32
(D) 2
+ 32
Sol. [1]
A1 = 21
01 x dx
A1 = 2 0 2
12t dt
A1 = 4 . 13
42 3
{k=skQy
51. ekuk a, b, c r Fkk d ' kwU; sr j l a[ ; k, ¡ gSA ; fn j s[ kkvksa4ax + 2ay + c = 0 r Fkk 5bx + 2by + d = 0 dki zfrPNsn fcanq pkSFks prqFkkZa' k esa gS rFkk nksuksa v{kksa l s l enwj LFkgS] r ks%(A) 2bc – 3ad = 0 (B) 2bc + 3ad = 0(C) 3bc – 2ad = 0 (D) 3bc + 2ad = 0
Sol. [3]4ax + 2ay + c = 05bx + 2by + d = 0ekuk i zfr PNsnh fcUnq (, – ) gSa4a – 2a + c = 0
= – c2a
....(1)
5b – 2b + d = 0
= – d3b
....(2)
– c d2a 3b
3bc 2ad3bc – 2ad = 0
52. ekuk PS , d f=kHkqt dh ekf/; dk gS ft l ds ' kh"kZ P(2, 2),Q(6, –1) r fkk R(7, 3) gSA (1, –1) l s gksdj t kus okyhj s[ kk] t ks PS ds l ekar j gS] dk l ehdj .k gS%(A) 4x – 7y – 11 = 0 (B) 2x + 9y + 7 = 0(C) 4x + 7y + 3 = 0 (D) 2x – 9y – 11 = 0
JEE MAIN Examination(2014) (Code - H)(Page # 20)
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Sol. [2]PS dk <ky
m = 2 1 1
13 922 2
m = 29
j s[ kk dh l ehdj .k
y + 1 = 29
(x – 1)
9y + 9 = 2x + 22x + 9y + 7 = 0
53. 2
2
0x x)xcossin(Lim
dk eku gS%
(A) 2
(B) 1
(C) – (D) Sol. [4]
2x 0
sin( c'x)limx
2
2 2x 0
sin( c x) (1 cx)lim (1 cx)( c x) x
= 1 . . 12
. 2
=
54. ; fn X = {4n – 3n – 1) : n N} r FkkY = {9 (n – 1) : n N} gS] t gk¡ N, i zkd r l a[ ; kvksadk l eqPp; gS] r ks X Y cj kcj gS%(A) N (B) Y – X(C) X (D) Y
Sol. [4]x : 4n – 3n –1x : (1 + 3)n – 3n – 1x : 1 + 3n + nC2 3
2 + nC3 33 + ........ –3n–1
x : 9 {nC2 + nC3 3 + ......} ...(1)r Fkk y : 9 (n – 1) ...(2)l ehdj .k (1) o (2) l s
9 dk xaq.kt vr % X Y Y
55. nh?kZoÙk x2 + 3y2 = 6 ds dsanz l s bl dh fdl h Li ' kZ j s[ kki j [ khaps x, yac ds i kn dk fcanq i Fk gS%(A) (x2 – y2)2 = 6x2 + 2y2
(B) (x2 – y2)2 = 6x2 – 2y2
(C) (x2 + y2)2 = 6x2 + 2y2
(D) (x2 + y2)2 = 6x2 – 2y2
Sol. [3]
nh?kZoÙk 2 2x y 16 2
; fn yEc dk i kn (h,k) gSa r cj s[ kk dh l ehdj .k xh + ky = k2 + h2
is y = hk
x + 2 2k h
k
vc ; fn ; g Li ' kZ j s[ kk gSa r c Li ' kZr k dk i zfr cU/kc2 = a2m2 + b2
2 2k hk
= 2h6
k
+ 2
22 2
2
k h
k
=
2
26h 2k
(k2 + h2)2 = 6h2 + 2k2
(x2 + y2)2 = 6x2 + 2y2
56. r hu /kukRed l a[ ; k, a c<+r h xqa.kksÙkj Js<+h esa gSA ; fn blxq.kksÙkj Js<+h dh chp okyh l a[ ; k nqxquh dj nh t k, ] t ks ubZcuh l a[ ; k, a l ekar j Js<+h esa gks t kr h gSA xq.kksÙkj Js<+h dkl koZvuqi kr gS%
(A) 2 + 3 (B) 3 + 2
(C) 2 – 3 (D) 2 + 3Sol. [4]
a, ar, ar2 xq.kksÙkj Js<h esa gSaa, 2ar, ar2 l ekUr j Js<h esa gSa
2ar = 2a ar
2
4r = 1 + r2
r2 – 4r + 1 = 0
r = 4 16 4
2
r = 4 2 3
2
r =2 3
bl fy, r = 2 + 3 xq.kksÙkj Js<h esa gksxk
JEE MAIN Examination(2014) (Code - H) (Page # 21)
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57. ; fn (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7
+..... ... + 10(11)9 = k(10)9 gS] r ks k cj kcj gs%
(A) 10121
(B) 100441
(C) 100 (D) 110Sol. [3]
109 + 2.111 108 + 3.112 107 ---- + 10.119
S = 109 (1+2 111
10
+ 3. 211
10
+ ..... + 10911
10
)
S111
10
= 109 (111
10
+ 2211
10
+ 3311
10
+.......101011
10
)
11s 110
=109 1 2 9 1 01 1 1 1 1 1 1 1
1 1 01 0 1 0 1 0 1 0
= 109
10
10111 1
1110 1011 10110
S
10 1091 11 1110 1 10 10
10 10 10
= 109 1010 1011 111
10 10
S . 1
10
= 109 (–10)
S = 100. 109
bl fy, K = 100
58. nks j s[ kk, ¡] ft uds fnd~&dksT; k] l ehdj .kksa1 + m + n = 0 r Fkk l2 = m2 + n2 dks l ar q"V dj r s gSa]ds chp dk dks.k gS :
(A) 3
(B) 4
(C) 6
(D) 2
Sol. [1]l + m + n = 0l2 – m2 – n2 = 0(m + n)2 – m2 – n2 = 0mn = 0
m = 0 ; k n = 01.l + 1.m + 1.n = 0, 1.l + 1.m + 1.n = 00.l + 1.m + 0.n = 0, 0.l + 0.m + 1.n = 0
l m n1 1 1 1 1 11 0 0 0 0 1
,
l m n1 1 1 1 1 10 1 1 0 0 0
l m n1 0 1
; k
l m n1 1 0
Cos = 1 0 02 2
Cos = 12
= 3
59. i j oy; ksa y2 = 4x r Fkk x2 = – 32y nksuksa dks Li ' kZ dj usokyh j s[ kk dh i zo.kr k gS :
(A) 21
(B) 23
(C) 81
(D) 32
Sol. [1]y2 = 4x, x2 = –32yLi ' kZ j s[ kk dh l ehdj .k] Li ' kZ j s[ kk dh l ehdj .k
y = mx + 1m
, y = mx + 8m2
1m
= 8m2
m = 12
60. ; fn x = – 1 r Fkk x = 2f(x) = log |x| = x2 + x ds pj efcanq gSa] r ks :(A) d (B) d(C) d (D) d
Sol. [3]
f'(x) = 2 x 1x
x i j = –1, 2 f'(x) = 0 – – 2+ 1 = 0 ....(1)
2
+ 4 + 1 = 0 .... (2)
= 2, = 12
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[CHEMISTRY]61. NO dkSu&l k fuEu xq.k i znf' kZr ugha dj r k gS\
(1) ; g vkWDl ht u l s ; ksx dj ukbVªkst u MkbZvkWDl kbMcukrk gSA(2) bl dh cU/k dksfV 2.5 gSA(3) xSl h; voLFkk esa i zfr pqEcdh; gSA(4) ; g , d mnkl hu vkWDl kbM gSA
Sol. 3(1) NO + 1/2 O2
NO2
(2) NO dk ca/k Øe 2.5 gSA(3) ; g xSl h; voLFkk esa vuqpqEcdh; gSA(4) ; g mnkl hu vkWDl kbM gSA
62. ; fn Z l ai hM+u xq.kd gks r ks de nkc i j okaMjokYl l ehdj .kdks fy[ kk t k l dr k gS %
(1) Z = 1 – RTPb
(2) Z = 1 + RTPb
(3) Z = 1 + PbRT
(4) Z = 1 – VRT
a
Sol. 4, d eksy xSl ds fy, okW.Mj okWy l ehdj .k ds vuql kj
2
aP (V b) RTV
fuEu nkc i j vk; r u mPp gksxkA b dks ux.; eku l dr s gSAvr %,
2
aP (V) RTV
aPV RTV
Z = PVRT
= 1 – a
VRT
63. /kkr q t ks vi us yo.kksa ds t yh; foy; uksa ds bySDVªkysfl l¼fo| qr vi ?kVu½ l s i zkIr ugha gks l dr h gS %(1) Cu(2) Cr(3) Ag(4) Ca
Sol. 4
22 2H o /H Ca / Ca
Eº Eº
Ca vf/kd fo| qr /kuh gS vr % t yh; foy; u i z; qDr ughagks l dr kA
64. , d oS| qr vi ?kV~; esa 0.2 M foy; u dk i zfr j ks/k 50 gSAbl foy; u dk fof' k"V pkydRo 1.4 S m–1 gSA bl h fo| qrvi ?kV~; ds 0.5 M foy; u dk i zfr j ks/k 280 gSA fo| qrvi ?kV~; ds 0.5 M foy; u dh eksyj pkydr k S m2
eksy –1 esa gksxh %(1) 5 × 103
(2) 5 × 102
(3) 5 × 10–4
(4) 5 × 10–3
Sol. 3
R = a
50 = 1
1.4 a
170ma
R = a
280 = 1 70
= 0.25 Sm–1
= 0.25 × 10–2 –1 - cm–1
m
1000c
=
–21000 0.25 100.5
= 5 –1-cm2 mole–1
= 5 × 10–4 S-m2 mol–1
65. CsCl dk; dsfUnzr ?kukdj t kyd esa fØLVfyr gksr k gSA
; fn fdukj s dh yEckbZ 'a' gks r ks fuEu l w=kksa esa l s dkSu&l kBhd gksxk\
(1) Cs Cl3r r a2
(2) Cs Clr r 3a
(3) Cs Clr r 3a
(4) Cs Cl3ar r2
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Sol. 1Cl– , SC cukr k gSACs+ ?ku ds dsUnz i j t kr k gSA
–3ar r2
66. 0.500 M C2H5OH(t y h; ), 0.100 MMg3(PO4)2(t yh; ), 0.250 M KBr(t yh; ) vkSj 0.125M Na3PO4(t yh; ) foy; uksa dks 25ºC i j /; ku nhft ; sAl Hkh uedksa dks i zcy bySDVªksykbV ekur s gq, fuEu dFkuksa esal s dkSu&l k dFku ; FkkFkZ gS\(1) 0.125 M Na3PO4(t yh; ) dk vkl ekfVd nkc
mPpr e gksxkA(2) 0.500 M C2H5OH(t yh; ) dk vkl ekfVd nkc
mPpr e gksxkA(3) bl l c ds fy, vkl ekfVd nkc ds eku l eku gksxkA
(4) 0.100 M Mg3(PO4)2(t yh; ) dk vkl ekfVd nkc
mPpr e gksxkASol. 3
i ic0.5 M C2H5OH (aq.) 1 0.50.10 M Mg3 (PO4)2 5 0.50.25 M KBr 2 0.50.125 M Na3PO4 4 0.5l Hkh l eku v.kql a[ ; d xq.k/keZ j [ kr s gSA
67. fuEu fdu vfHkfØ; kvksa esa H2O2 , d vi pk; d dk dke
dj r k gS\(a) H2O2 + 2H+ + 2e– 2H2O(b) H2O2 – 2e– O2 + 2H+
(c) H2O2 + 2e– 2OH–
(d) H2O2 + 2OH– –2e– O2 + 2H2O(1) (a), (c)(2) (b), (d)(3) (a), (b)(4) (c), (d)
Sol. 2D; ksafd vi pk; d gS
01 –
22 2H O O 2e
(b) & (d)
68. ; kSfxdksa CH3Cl, CH3CH2Cl, (CH3)2CHCl vkSj(CH3)3CCl dk SN2 fØ; k esa fØ; k dj .k dk mfpr Lr j
Øe gksr k gS %(1) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl(2) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl(3) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl(4) CH3Cl> CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl
Sol. 4SN2
SN2 dh fØ; k' khyr k 1
f=kfoe ck/kkSN2 dh vksj fØ; k' khyr k dk Øe gSCH3Cl > CH3–CH2–Cl > (CH3)2CH–Cl > (CH3)3C–Cl
69. M3+ /kkrq vk; u dk pkj , d i dM+h fyxSaMksa] L1, L2, L3 vkSj
L4 ds l kFk v"V Qydh; l adj yky] gj s] i hys vkSj uhys
LFkyksa l s r j axnS/; kZsa dk Øekuql kj vo' kks"k.k dj r k gSA pkj
fyxSaM+ksa dh ' kfDr dk c<+r k Øe gS %(1) L3 < L2 < L4 < L1
(2) L1 < L2 < L4 < L3
(3) L4 < L3 < L2< L1
(4) L1 < L3 < L2 < L4
Sol. 4
(a) Li sDVªks j l k; fud Js.kh ds vuql kj vf/kd vo' kks"k.k
vkofÙk gksus i j fyxs.M i zcy gksxkA
V I B G Y O R
EL1 < L3 < L2 < L4
(yky) (gj k) (i hyk) (uhyk)
70. ukbVªkst u ds vkdyu ds fy, 1.4 xzke dkcZfud ; kSfxd
t sYMkWy fof/k ds vuql kj vi fpr fd; k x; k r Fkk eqDr gq,
veksfu; k dks 60 feyh 10M
l Y¶; wfj d vEy esa vo' kksf"kr
fd; k x; kA vf/k' ks"k vEy ds i w.kZ mnkl huhdj .k ds fy, 20
feyh 10M
l ksfM; e gkbMªkWDl kbM dh vko' ; dr k gqbZA ; kSfxd
esa ukbVªkst u dh i zfr ' kr r k gS %(1) 3 % (2) 5 %(3) 6 % (4) 10 %
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Sol. 41.4 gm NH3
2 4M M60mL H SO 20mL NaOH10 10
ekuk x % N
x1.4 1 20 1 60100 1 214 10 1000 10 1000
14 100x (0.012 – 0.002)1.4
= 10 %
71. l kUnz.k C i j vkSj vuUr r uqr k i j NaCl foy; u dh
bfDooySUV pkydrk dks C vkSj ekur s gq, mudk vki l h
l EcU/k fy; k t k l dr k gS %
(B , d fLFkj vad gS)
(1) C = – (B) C
(2) C = + (B) C
(3) C = + (B) C
(4) C = – (B) C
Sol. 1
c = — CB
72. vfHkfØ; k SO2(g) + 21
O2(g) SO3(g), ds fy, Kp =
KC (RT)x gksxk t cdh l c l wpd v{kj l kekU; vFkZ j [ kr s
gSa r ks vkn' kZj : i r k ekur s gq, x dk eku gksxk %(1) 1/2(2) 1(3) –1(4) – 1/2
Sol. 4
SO2(g) 21 O (g)2
SO3(g)
Kp = Kc (RT)ng
ng = 1 – (1 + 21
) = 21–
Kp = Kc (RT)–1/2
73. vfHkfØ; k l sV]
CH3COOH LiAlH4 A PCl5 B Alc. KOH C,
esa fØ; k Qy C gksr k gS %
(1) bFkkbZyhu
(2) , fl Vkby Dyksj kbM
(3) , sfl V, sfYMgkbM
(4) , fl fVyhuSol. 1
CH3–CH2–OH
CH3–CH2–Cl
CH2=CH2 (bFkkbZfyu)
74. l ksfM; e QSukDl kbM dh mPp nkc vkSj 125°C i j CO2 l s
vfHkfØ; k djus i j t ks ; kSfxd i zkIr gksrk gS ml ds , fl fVys' ku
i j fØ; k Qy C gksr k gSA
ONa
+ CO2 B C125º H+
5 Atm Ac O2
cM+h ek=kk esa fØ; k Qy C gksxk (
(1) COOCH3
OH
(2)
OCOCH3
COOH
(3) COOH
OCOCH3
(4)
OH
COCH3
COCH3
Sol. 3
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75. , sfyQSfVd i zk; ej h , ehu dks Dyksj ksQkeZ vkSj , Fkkuksfyd
i ksVSf' k; e gkbMªkDl kbM ds l kFk xje djus i j cuk vkj xSfud; kSfxd gksr k gS %
(1) , d , sfYdy fl ; kukbM(2) , d , sfYdy vkbl ksfl ; kukbM
(3) , d , sYdkuksy(4) , d , sYdsuMk; ksy
Sol. 2
R—NH2 + CHCl3 +KOH R–NC
1º-, ehu Dyksj ksQkWeZ , fYdy vkbl ks l k; ukbM; g dkfcZy , ehu vfHkfØ; k ; k vkbl ks l k; ukbM i j h{k.k gSA
76. CsI3 v.kq ds fy, ; FkkFkZ dFku gksxk %(1) bl esa Cs3+ vkSj I– vk; u gksr s gSA
(2) bl esa Cs+, I– vkSj I2 t kyd gksr s gSA
(3) ; g , d l gl a; kst dh v.kq gSA(4) bl esa Cs+ vkSj I3
– vk; u gksr s gSASol. 4
CsI3 Cs+ + I3–
77. l ehdj .k t ks l ar qfyr gS vkSj ; FkkFkZ fØ; k Qyksa dh l wpd gS]
gSa %
(1)[Mg(H2O)6]2+ + (EDTA)4– NaOH dk vkf/kD;
[Mg(EDTA)]2+ + 6H2O(2) CuSO4 + 4KCN K2[Cu(CN)4] + K2SO4
(3) Li2O + 2KCl 2LiCl + K2O(4) [CoCl(NH3)5]
+ + 5H+ CO2++5NH4++ Cl–
Sol. 1; g vfHkfØ; k dBksj t y esa Mg2+ dk i r k yxkus ds fy,i z; qDr gksr h gSA
78. fuEu esa l s fdl v.kq ds fy; s cgqr l hek r d 0 gksxk?
(a)
Cl
Cl
(b)
CN
CN
(c)
OH
OH
(d)
SH
SH(1) dsoy (c)(2) (c) vkSj (d)(3) dsoy (a)(4) (a) vkSj (b)
Sol. 2
(a) (b)
(c) (d)
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79. j l k; fudr k fj Dr vfHkfØ; k 2A + B C + D, esa r hui Fkd i z; ksxksa esa 298 K i j fuEu xfr d vkadM+s i zkIr fd; sx; sA
3
3
3
104.2M1.0M2.0102.1M2.0M1.0102.1M1.0M1.0
)SLmol(
CofformationofrateInitial
)B(ionConcentrat
Initial
)A(ionConcentrat
Initial
i zkj fEHkd l kanz.k
i zkj fEHkd l kanz.k
cuus dh i zkj fEHkdnj ¼eksy ½(A) (B)
CL S– –
vfHkfØ; k ds fy; s C cuus dk nj fu; e gksxkA
(1) dtdc
= k[A] [B]2
(2) dtdc
= k[A]
(3) dtdc
= k[A] [B]
(4) dtdc
= k[A]2 [B]
Sol. 22A + B C + D(1) & (3) [A]1
(1) & (2) [B]º r = k[A]1
80. buesa l s vfHkfØ; kvksa dk dkSu&l k Øe ; FkkFkZ : i esa yksgs vkSj
bl ds ; kSfxdksa dh j k; k; fud vfHkfØ; kvksa dks fu: fi r dj r kgS\
(1) Fe Cl ,heat2 FeCl3 heat, air FeCl2
Zn Fe
(2) Fe O heat2, Fe3O4 CO, 600ºC FeO
CO, 700ºC Fe
(3) Fe dil H SO2 4 FeSO4 H SO2 4, 2O Fe2(SO4)3
heat Fe
(4) Fe O heat2, FeO dil H SO2 4 FeSO4
heat FeSol. 2
Fe Heat
O2 Fe3O4 Cº600
CO FeO Cº700
CO Fe
(FeO + Fe2O3); g vfHkfØ; k okR; k Hkêh esa i z; qDr gksr h gSA
81. t yh; foy; u esa , ehuksa dh {kkj h; i zofÙk ds vuql kjfuEufyf[ kr ksa esa l s fdl ds fy, pKb dk eku de l s de
gksxk\(1) (CH3)3N(2) C6H5NH2(3) (CH3)2NH(4) CH3NH2
Sol. 3
foLFkkfud r , dkdh ; qXe bysDVªkWu
(U; wur e {kkj h; )t yh; foy; u esa {kkj h; l keF; Z dk Øe
{kkj h; l keF; Z Kb bpk
1
(CH3)2NH vf/kd {kkj h; gSA vr % ; g U; wur e pKb j [ kr kgSA
82. fuEu {kkj ksa esa l s dkSu , d DNA esa ugha i k; k t kr k?(1) l kbVksl hu(2) FkkbZehu(3) D; wuksyhu(4) , fMuhu
Sol. 3
(D; wuksyhu)
D; wuksyhu DNA esa mi fLFkr ugha gksr k gSAl kbVksl hu] FkkbZehu rFkk , fMuhu DNA esa mi fLFkr gksrss gSA
83. : fcfM; e i j ek.kq (Z = 37) ds fy; s l a; kst h bySDVªkWuksa dsmfpr pkj DokUVe uEcj ksa dk l sV gksr k gS ?(1) 5, 1, 1 + 1/2(2) 5, 0, 1, + 1/2(3) 5, 0, 0, + 1/2(4) 5, 1, 0, + 1/2
Sol. 3
84. 1,1,1-VªkbDyksj ksbZFksu dks fl Yoj i kmMj ds l kFk fØ; kdjus i j l cl s cM+h ek=kk esa cuk vkj xSfud ; kSfxd gksr k gSA?(1) 2 C; qVkbu(2) 2 - C; qVhu(3) , fl fVyhu(4) bZFkhu
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Sol. 1
CH3—CC—CH3 + 6AgCl2-butyne
85. uhps dqN v) Z l sy vfHkfØ; k, nh xbZ gSAMn2+ + 2e– Mn : Eº = – 1.18 V2(Mn3 + e– Mn2+) : Eº = + 1.51 V3 Mn2+ Mn + 2Mn3+ ds fy; s Eº gksxkA(1) – 0.33 V ; vfHkfØ; k ugh gksxh(2) – 0.33 V ; vfHkfØ; k gksxh(3) – 2.69 V ; vfHkfØ; k ugh gksxh(4) – 2.69 V ; vfHkfØ; k gksxh
Sol. 3Mn2+ + 2e– Mn ; Eº = – 1.18 V2Mn3+ + 2e– 2Mn2+ ; Eº = +1.51 V(1) – (2)Mn2+ + 2e– + 2Mn2+ 2Mn3+ + 2e–1 + MnEº = – 1.18 – 1.51 ( n l eku gS) Eº = – 2.69 V(3) vfHkfØ; k i zkIr ugha gksxh gSA
86. , d fo' ks"k xSl h; feJ.k esa vkWDl ht u vkSj ukbVªkst u ds
nzO; ekuksa dk vuqi kr % 1 : 4 gSa A feJ.k esa budh v.kql a[ ; kvksa dk vuqi kr gksxkA(A) 1 : 8 (B) 3 : 16(C) 1 : 4 (D) 7 : 32
Sol. 4O2 : N2
nzO; eku 1 : 4
eksy321
:284
7 : 32
87. buesa l s fdl dks dUMSUl s' ku cgqyd ekuk t k; sxk ?(A) VSQykWu (B) , sfØyksukbVªkby
(C) MSdjku (D) fu; ksfi zuSol. 3
Vsfj yhu ; k MsØksu
–nH O2 ( )l a?kuu
(Msdj ku)
Msdj ku ds vykok vU; l Hkh ; ksxkRed cgqyd gSA
88. fuEu vkDl ks vEyksa ds fy, vEy ' kfDr dk ; FkkFkZ ?kVrk Øegksxk %(A) HClO4 > HClO3 > HClO2 > HOCl(B) HClO2 > HClO4 > HClO3 > HOCl(C) HOCl > HClO2 > HClO3 > HClO4(D) HClO4 > HOCl > HClO2 > HClO3
Sol. 1ClO4
– (vf/kd vuqukn LFkk; h )
89. , FksukWy ds i w.kZ Toyu ds fy; sC2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l),ce dsyksj hehVj esa ekfi r Åt kZ 25ºC i j 1364.47 kJmol–1 gSA vkn' kZr k ekur s gq, Toyu dh , UFkSYi h CH,gksxh A(R = 8.314 kJ mol–1)(A) –1460.50 kJ mol–1
(B) –1350.50 kJ mol–1
(C) –1366.95 kJ mol–1
(D) –1361.95 kJ mol–1
Sol. 3H = U + ngRT= – 1364.47 + (–1) × 8.314 × 10–3 × 298= –1364.47 – 2.47= – 1366.95 kJ mol–1
90. R – CH2 – OH R – CHO esa cnyus dk l cl s vf/kd
mi ; qDr vfHkdkj d gksr k gSA(A) CrO3
(B) PCC (fi fj Mhfu; e Dyksj ksØksesV)(C) KMnO4(D) K2Cr2O7
Sol. 2
R—CH2—OH
PCC (fi j hfMfu; e Dyksj ks ØksesV) nqcZy vkWDl hdkjd gSA
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