JEE MAIN EXAMINATION - 2015 QUESTION WITH SOLUTION PAPER CODE - A€¦ · (Page # 2) JEE MAIN...
Transcript of JEE MAIN EXAMINATION - 2015 QUESTION WITH SOLUTION PAPER CODE - A€¦ · (Page # 2) JEE MAIN...
JEE MAIN EXAMINATION - 2015
QUESTION WITH SOLUTION
PAPER CODE - A
[HINDI VERSION]
Fastest Growing Institute of Kota (Raj.)
FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | CBSE | SAT | NTSE | OLYMPIADS
JEE MAIN Examination(2015) (Code - A)(Page # 2)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
[PHYSICS]1. nks i RFkj 240 m Å¡ph , d pV~Vku ds fdukj s l s , d l kFk
Åi j dh vksj Øe' k% 10 m/s o 40 m/s dh i zkjafHkd pkyl s Qsads t kr s gSaaA i zFke i RFkj ds l ki s{k f} r h; i RFkj dhl ki sf{kd fLFkfr dk l e; i fj or Zu dks fuEu esa l s fdl xzkQesa mfpr cr k; k x; k gS \(ekuk i RFkj /kj kr y i j Vdj kus ds ckn mNyr k ugha gS r Fkkok; q i zfr j ks/k ux.; ] g = 10 m/s2 gSA)(fp=k O; ofLFkr gS r Fkk i Sekuk ughas fy; k gSA)
(1)
8t 12t(s)
(y –y ) m2 1
240
(2)
8 12t(s)
(y –y ) m2 1
240
(3)
8 12t(s)
(y –y ) m2 1
240
(4)
12 t(s)
(y –y ) m2 1240
Sol. 3i zFke i RFkj
0 t 8secvr = 40 – 10= 30 m/sar = 0
t=2sec
t=6sec
sr = vr × t = 30 × 8 = 240 m
(y –y )m2 1
240
8t(sec)
`
t=8sec
t=1sec
8 sec < t 12 secvr i fj ek.k esa c<+r k gS r Fkk l ki sf{kd Roj .k g uhps dh vksj
(y –y )m2 1
240
8 12t(sec)
2. , d l j y yksyd dk nksyudky LT 2g gSA L dk
eki k x; k eku 20.0 cm gS] 1 mm ' kq) r k gS r Fkk yksyd
ds 100 nksyuksa ds fy, 1s fo; kst u {ker k dh , d ?kM+h l s
90 s eas i zkIr gksr s gSaA g dh ' kq) r k Kkr gksxh :(1) 1% (2) 3%(3) 2% (4) 5%
Sol. 2
dT 1 dL 1 dgT 2 L 2 g
90 1100 100
1 dg 1 dL dT2 g 2 L T
1 0.1 1/100 1 12 20 90 /100 400 90
1 dg 1 12 g 400 90
dg 490 2g 400 90
490200 90 = 0.02722
3%2.72%100g
dg
JEE MAIN Examination(2015) (Code - A) (Page # 3)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
3.
A BF
fn; s x; s fp=k esa nks xqVds A vkSj B ft uds Hkkj Øe' k%
20 N o 100 N gSA ; s fp=kkuql kj , d cy F ds } kj k nhokj
ds fo: ) nck; s x; s gSaA ; fn xqVds ds chp ?k"kZ.k xq.kakd 0.1
gS r Fkk xqVds B o nhokj ds chp 0.15 gS] xqVds B i j nhokj
ds vkj ksfi r ?k"kZ.k cy gksxk :(1) 100 N (2) 150 N(3) 120 N (4) 80 N
Sol. 3
A BF
ekuk fd fudk; l kE; voLFkk esa gS] dqy xq: Rokd"kZ.k cynhokj l s ?k"kZ.k cy ds } kj k l Ur qfyr gksxkA?k"kZ.k cy = 120 N
4. nzO; eku m dk , d d.k pky 2l s x fn' kk esa xfr ' khy gSt ks nwl j s d.k ft l dk nzO; eku 2m dh pky l s y fn' kk esa
xfr ' khy gS l s Vdj kr k gSA ; fn VDdj i w.kZr % vi zR; kLFk gS]VDdj ds nkSj ku Åt kZ esa i zfr ' kr gkfu yxHkx gS %
(1) 44% (2) 62%(3) 56% (4) 50%
Sol. 3
m 2v
v
2m
VDdj ds i gys
xP
= 2mv i
yP
= 2mv j
V
3m
VDdj ds ckn
xP
= 3mv'cos
yP
= 3 v' sin
l aosx l aj {k.k l s %
{kSfr t esa 2mv = 3mv' cos ...(i)
m/okZ/kj esa 2mv = 3mv' sin ...(ii)
(i) vkSj (ii) l s tan = 1; = 45°
vfUr e pky v' = 3
v22
i zkj fEHkd K.E. ; 1/2 (m) (2v)2 + 1/2 (2m) (v)2 = 3mv2
vfUr e K.E. ; 1/2 (3m)
2
3v22
= 4/3 mv2
% deh i
fi
)KE()KE()KE(
× 100%
= 55.55 ~ 56%
5. , d Bksl , d l eku ' kadq ds ' kh"kZ ds nzO; eku dsUnz dh nwj h
z0 gSA ; fn bl ds vk/kkj dh f=kT; k R gS r Fkk bl dh Å¡pkbZ
h gS] r c z0 cj kcj gS &
(1) 5h8 (2)
3h4
(3) 5h8
(4) 23h
8R
Sol. 2
03hz4 z0
h
(theory l s)
JEE MAIN Examination(2015) (Code - A)(Page # 4)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. 2Bksl xksyk ft l dk nzO; eku M, f=kT; k R gSAxksyh; Hkkx gVk; k t kr k gS ft l dh f=kT; k R/2, bl fy,bl dk nzO; eku M/8
xqgk ds dsUnz i j foHko= Vsolid sphere + Vremoved part
CR/2
=
3
GM2R
22 R3R
2 + )2/R(2)8/M(G3
= RGM
8. , d yksyd , d l eku r kj ft l dk vuqi zLFk dkV {ks=kQy A
dk cuk gS ft l dk vkor Zdky T gSA t c , d vfr fj Dr
nzO; eku M bl xsan esa feyk; k t kr k gSA r ks vkor Zdky
cnydj TM gSA ; fn r kj ds i nkFkZ dk ; ax xq.kkad Y gS] r c
1/Y cj kcj gS & (g = xq: Roh; Roj .k)
(1)
2MT A1T Mg (2)
2
M
T A1T Mg
(3)
2MT A1T Mg (4)
2MT Mg1T A
Sol. 1
T = 2 g
; TM = 2 g'
=
Mg/ A/
' =
MgA =
' = 1 +
MgA
Also:
TTM =
' TM =
1/2MgT 1A
2
2M
TT
= 1 + AMg
1
TT
2
2M
= MgA
1
= A
Mg
1TT 2
M
6. nzO; eku M o f=kT; k R ds , d Bksl xksys l s vf/kdr e l EHko
vk; r u dk ?ku dkVk x; k gSA ?ku ds dsUnz o bl ds , d
Qyd ds yEcor ~ xqt j us okyh v{k ds i fj r % ?ku dk t M+Ro
vk?kw.kZ gksxk %
(1)
2MR32 2
(2)
24MR3 3
(3)
24MR9 3 (4)
2MR16 2
Sol. 3
I = 6
Mx2
fdukj s dh yEckbZ : (x)
2R = 3 x (fod.kZ)
x = 3R2
vc]?ku dk nzO; eku :
m =
3R34
M
3
3R2
R
R
3R4M3
33R8 3
m = 3
M2
I = 31
3M2
3R4 2
= 39
MR4 2
7. fp=kkuql kj nzO; eku M o f=kT; k R dk , d Bksl xksys l sf=kT; k R/2 dk xksyh; Hkkx gVk; k x; k gSA xq#Roh; foHkoV = 0, r = i j ekus] cuh xqgk ds dsUnz i j foHko gksxk(G = xq#Roh; fu; r kad)
(1) R3GM2
(2) RGM
(3) R2
GM(4)
RGM2
JEE MAIN Examination(2015) (Code - A) (Page # 5)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
9. r ki T i j f=kT; k R ds , d xksyh; dks' k i j fopkj dhft , Abl ds vUnj d f".kdk fofdj .k QksVksuksa dh vkn' kZ xSl dsvuql kj eku l dr s gSa] ft l dh vkUr fj d Åt kZ i zfr , dkad
vk; r u 4Uu TV
r Fkk nkc
1 Up3 V gSA ; fn
dks' k vc , d : ) ks"e i zl kj ds vUrxZr gS r ks T o R ds chpl EcU/k gksxk :
(1) RT e (2) 3
1TR
(3) 1TR
(4) 3RT eSol. 3
4Uu TV
1 UP3 V
: ) ks"e i zl kjTV–1 = K
4TV C
14
3 14
43
(r = )
4TV C
13TV C
1334T R C
3 1T
R
10. fu; e Å"ek {ker k 1 J/°C dh , d Bksl oLr q nks r j g ds
fj t oZj l s l Ei dZ esa j [ kus ds } kj k Åf"er gSA
(i) 2 fj t oZj ds l kFk mÙkj ksÙkj l Ei dZ esa j [ kus i j r kfd
i zR; sd i j Å"ek dh l eku ek=kk l s l IykbZ dj r k gSA
(ii) 8 fj t oZj ds l kFk mÙkj ksÙkj l Ei dZ esa j [ kus i j r kfd
i zR; sd fj t oZj Å"ek dh l eku ek=kk l IykbZ dj r k gSA
nksukas fLFkfr ; ksa esa oLr q i zkj afHkd r ki 100°C l s vfUr e r ki
200°C r d yk; k t kr k gSA nksuksa fLFkfr ; ksa esa oLr q dh
, UVªkWi h i fj or Zu Øe' k% gS %(1) ln2, 2ln2 (2) ln2, ln2(3) ln2, 4ln2 (4) 2ln2, 8ln2
Sol. 2
(i) 150 200
1100 150
dQ dT dTs ms msT T T
150 200ln ln100 150
3 4ln ln2 3
1s ln2
(ii) 112.5 125
2100 112.5
dQ dQ dQs ......T T T
112.5 125ln ln .....100 112.5
9 10 16ln ln ln8 9 15
16ln8 = ln2
11. , d foyfxr cUn psEcj esa , d vkn' kZ xSl i j fopkj dhft , AxSl , d : ) ks"e i zl kj ds vUr xZr ] v.kqvksa ds chp VDdj dkvkSl r pky Vq ds vuql kj c<+r k gS] t gk¡ V xSl dk vk; r u
gSA q dk eku gksxk
V
P
CC
:
(1) 6
53 (2)
21
(3) 2
1(4)
653
Sol. 3ek/; eqDr i Fk
2
12 d n
n v.kqvk sadh l a[ ; k
vk; r u
avg.v T T.V-1 = C
avg.
Vtv T v vk; r u gSA
1
2
r 1
V VC
v
1
q 2v v 1q2
JEE MAIN Examination(2015) (Code - A)(Page # 6)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
12. , d l j y yksyd ds fy, ] , d xzkQ bl ds foLFkki u ds l ki s{kbl dh xfr t Åt kZ (KE) o fLFkfr t Åt kZ ds (PE) ds chpvkj sf[ kr gSA fuEu esa l s dkSul k l gh gS\ (xzkQ O; ofLFkr gSr Fkk i Sekus l s ugha cus)
(1) d
KEE
PE
(2) d
PEE
KE
(3) d
KEE
PE
(4)
E
PE
KE
Sol. 2 KEmax ek/; fLFkfr i j
U=0KE=0
pj e fLFkfr pj e fLFkfrMPV=max=0
U=0KE=0
PEmin ek/; fLFkfr i j
13. , d j syxkM+h , d l h/ks i Fk i j 20 ms–1 dh pky l s xfr ' khygSA ; g 1000 Hz dh vkofÙk i j l hVh ct k j gh gSA i Fk dsfudV [ kM+s , d O; fDr ds } kj k ¼t c j syxkM+h ml l s xqt j r hgSA½ l quh vkofÙk esa i zfr ' kr i fj or Zu gksxk &(/ofu dh pky = 320 ms–1)(1) 18% (2) 12%(3) 6% (4) 24%
Sol. 2
f1 = 1000 320
300 20
= 1066 Hz
f2 = 1000 320
300 20
= 941 Hz
Change is ~ 12%
14. , d yEch csyukdkj dks' k esa Åi j h v) Z esa /kukRed i "Bvkos' k o fupyh v) Z esa _ .kkRed i "B vkos' k – gSAcsyu ds pkj ksa vksj fo| qr {ks=k j s[ kk, ¡ gSA fn; s x; s fp=k esafn[ kkbZ gS : (fp=k esa O; ofLFkr gS r Fkk i Sekuk ugha fy; k gSA)
(1) ++
+++++++
–––––––––
– (2) ++
+++++++
––––––––––
(3)
++
+++++++
––––––––––
(4)
++
+++++++
–––––––––
–
Sol. 1Tangent to the electrical field lines will giveus the direction at a given point.
15. f=kT; k R dk , d l eku : i l s vkosf' kr Bksl xksys dk foHko
V0 (ds l ki s{k eki k x; k gSA) bl dh l r g i j gSA bl xksys
ds fy, foHko 2V3 0 ,
4V5 0 ,
4V3 0 r Fkk
4V0 ds l kFk
l efoHko l r gas gS dh f=kT; k Øe' k% R1, R2, R3 r Fkk R4 gS]
r c %
(1) R1 = 0 r Fkk R2 > (R4 – R3)
(2) R1 0 r Fkk (R2 – R1) > (R4 – R3)
(3) R1 = 0 r Fkk R2 < (R4 – R3)
(4) 2R < R4
JEE MAIN Examination(2015) (Code - A) (Page # 7)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. 3 & 4
3V /20
V0
r=R
0 0 0 01 2 3 4
3V 5V 3V VR ; R ; R ; R
2 4 4 4
2 23
KQr R V 3R r2R
01
3Vv , R 0
2
2 2023
5V KQ 3R R4 2R
2RR2
r > R
0
3
3V KQ4 R
30
4KQ KQ R RR3V 3 KQ 3
0
4
V KQ4 R
40
4KQ 4KQR R 4RV KQ
r qyuk dj us i j i zkIr dj r s gS %(1) ; k (2)
16. fn; s x; s i fj i Fk esa] 2µF l a/kkfj =k i j vkos' k Q2 ,C dsvuql kj cnyr k gSA 1µF l s 3µF r d i fj ofr Zr gSA 'C' dsQyu ds : i esa Q2 mfpr fn; k x; k gS % (fp=k O; ofLFkrgS r Fkk i Sekuk ugha fy; k gSA)
C
1µF
2µFE
(1) Q2
1µF 3µFC
Charge
(2) Q2
1µF 3µFC
Charge
(3) Q2
1µF 3µFC
Charge
(4) Q2
1µF 3µFC
Charge
Sol. 2
3Cq EC 3 q = CV
q C
23C 2q E
C 3 3
22Cq E
C 3
22Cq E
31C
q = CV
C q2 ; fn C , q = fu; r eku
17. t c 5V foHkokUr j 0.1 m yEckbZ ds r kj i j vkj ksfi r fd; kx; k gS] r c bysDVªkWuksa dh vuqxeu pky 2.5×10–4 ms–1
gSA ; fn r kj esa bysDVªkWu ?kuRo 8×1028 m–3 gS] r ks i nkFkZdh i zfr j ks/kdr k yxHkx gS &(1) 1.6 × 10–6 m (2) 1.6 × 10–7 m(3) 1.6 × 10–8 m (4) 1.6 × 10–5 m
Sol. 4i = neAVd
RV
= neAVd { lRA
}
V A = neAVd
50.1 = 8 × 1028 × 1.6 × 10–19 × 2.5 × 10–4
= 1.56 × 10–5 m ~ 1.6 × 10–5 m
JEE MAIN Examination(2015) (Code - A)(Page # 8)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
18. i znf' kZr i fj i Fk esa] 1i zfr j ks/kd esa /kkj k gksxh &
(1) 1.3 A, P l s Q(2) 0.13 A, P l s Q(3) 0.13 A, Q l s P(4) 0A
Sol. 36V P0
9V
Qx
59x
+ 3
6x +
1x
= 0
15
x1530x527x3 = 0
x = 233
A
from Q to P
19. fHkUu&fHkUu f=kT; kvksa dh nks l ek{kh; i fj ukfydkvksa esa /kkj k I
l eku fn' kk esa i zokfgr gSA ekuk 1F
ckgjh ds dkj .k vkUr fj d
i fj ukfydk i j pqEcdh; cy gS rFkk 2F
vkUr fj d ds dkj .k
ckgj h i fj ukfydk i j pqEcdh; cy gks] r c %
(1) 1F
= 2F
= 0
(2) 1F
f=kT; h; : i l s ckgj dh vksj gS r Fkk 2F
= 0
(3) 1F
f=kT; h; : i l s vUnj dh vksj gS r Fkk 2F
= 0
(4) 1F
f=kT; h; : i l s vUnj dh vksj rFkk 2F
f=kT; h; : i
l s ckgj dh vksjSol. 1
vuqi zLFk dkV n' ; %
i2i1
(nksuksa i fj ukfydk, ¡ i zd fr esa vkn' kZ gSA )nksukas r kj i j Li j vkdf"kZr gksaxs] fdUr q dqy cy i zR; sd r kji j ' kwU; gksxkA
fl ) kUr %nks /kkj kokgh ?kVd i j Li j vkdf"kZr gSA ; fn /kkj k dh fn' kkl eku gSA
F.B.D
0F1
0F2
20. nks yEcs /kkj kokgh i r ys r kj ] nksuksa esa /kkj k I gS] t ks yEckbZ L
ds vpkyd /kkxsa ds } kj k LFkkfi r gS rFkk fp=kkuql kj l kE; koLFkk
esa gS] /kkxs Å/okZ/kj l s '' dks.k cukr s gSA ; fn r kj ksa dk
nzO; eku i zfr , dkad yEckbZ gS] r c I dk eku gksxk &
(g = xq: Roh; Roj .k)
(1)
0
gL2 tan
(2)
0
gL2sincos
L
II
(3)
0
gLsincos
(4)
0
gL tan
Sol. 2
xid
LL
d2i20
cy i zfr , dakd yEckbZ fy; k x; k gS
tan =
.g.d2i20
JEE MAIN Examination(2015) (Code - A) (Page # 9)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
i2 =
cos.sin.g.
0 (2) d [d = 2L sin]
i = 2 sin
cosLg
0
21. Hkqt k, ¡ 10 cm o 5 cm dk , d vk; r h; ywi esa 12 A dh
/kkj k I i zokfgr gSA fp=kkuql kj fofHkUu foU; kl ksa esa fLFkr gSA
(a)
x
I
I
I
I
z
y
B
(b)
x II
II
z
y
B
(c)
x
II
I
I
z
y
B
(d)
x II
II
z
y
B
; fn ; gk¡ i j , d l eku pqEcdh; {ks=k 0.3 T /kukRed zfn' kk esa] foU; kl ksa esa ywi (i) LFkk; h l kE; koLFkk r Fkk (ii)vLFkk; h l kE; koLFkk esa gksxk \(1) Øe' k% (a) r Fkk (b)(2) Øe' k% (b) r Fkk (c)(3) Øe' k% (b) r Fkk (d)(4) Øe' k% (a) r Fkk (c)
Sol. 3l kE; voLFkk ds fy,
= 0
= MB sin n; fn, sin = 0;
= 0
; fn M
o B
ds chp dks.k ' kwU; gS r c LFkk; h l kE; voLFkk
; fn M
o B ds chp dks.k gS\ r c vLFkk; h l kE; voLFkk
22. , d i zsj d Ro (L=0.03H) r Fkk , d i zfr j ks/kd(R=0.15 k) fp=kkuql kj i fj i Fk esa 15V EMF dh cSVªhdh Js.khØe esa l a; ksft r gSA dqat h K1 yEcs l e; ds fy,yxk; h t kr h gSA r c t = 0 i j , K1 [ kqyh gS r Fkk dqat h K2
l kFk&l kFk yxh gSA t=1 ms i j ] i fj i Fk esa /kkj k gksxh &
5(e 150)
(1) 6.7 mA (2) 67 mA(3) 100 mA (4) 0.67 mA
Sol. 4nh xbZ fLFkfr ds vuql kj :
i0 = RV
= 31015.015
= 0.1A
i = i0 LRt
e
= 0.1 × 03.0101015.0 33
e
= 0.1 × e–5 = 150
1.0 = 0.67 mA
23. , d yky LED bl ds pkj ksa vksj , d l eku : i l s 0.1watt dk i zdk' k mRl ft Zr dj r h gSA Mk; ksM l s 1 m dh nwjhi j i zdk' k ds fo| qr {ks=k dk vk; ke gksxk &(1) 5.48 V/m (2) 2.45 V/m(3) 1.73 V/m (4) 7.75 V/m
Sol. 2
r hozr k 20
P 1 E CA 2
202
P 1 E C24 R
20
2PE4 R C
9
8
2 0.1 9 10E1 1 3 10
9
8
1.8 10 18E33 10
E 6 2.45 V /m
24. , d o.khZ; i zdk' k dks.k A dh , d dkap fi zTe i j vki fr rgSA ; fn fi zTe ds i nkFkZ dk vi or Zukad gS] , d fdj .kQyd AB i j dks.k i j vki fr r gS t ks i z; qDr fi zTe dsQyd AC l s i kj xfer gksxh %
A
B C
(1) > sin–1 1 1sin A sin
(2) < cos–1
1sinAsin 1
(3) > cos–1
1sinAsin 1
(4) > sin–1 1 1sin A sin
JEE MAIN Examination(2015) (Code - A)(Page # 10)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. 1r2 < cr2 < sin–1 (1/)
A
r1 r2
sin r2 < 1/sin = sin r1r1 = sin–1 (sin/)sin (A – r1) < 1/
sin
sinsinA 1
< 1
A – sin–1
sin
< sin–1
1
A – sin–1
1
< sin–1
sin
m1sinAsin 1
< sin
sin–1
m1sinAsin 1
<
25. xehZ dh j kr ] ok; q dk vi or Zukad /kj kr y ds fudV U; wur e
gS r Fkk /kj kr y l s Å¡pkbZ ds l kFk c<+r k gSA t c , d i zdk' k
i qUt {kSfr t : i l s gS] gkbxsu ds fl ) kUr l fEefyr gSA bl ds
xeu dj us i j ] i zdk' k i qUt %
(1) uhps dh vksj eqM+r h gSA
(2) fcuk fdl h fo{ksi .k ds {kSfr t t kr h gSA
(3) l dM+h gks t kr h gSA
(4) Åi j dh vksj eqM+r h gSASol. 4
Åi j dh vksj eqM+r h gSA
26. ekuk O; fDr dh i qr yh dh f=kT; k 0.25 cm r Fkk 25 cm
nwj ns[ kus ds fy, vko' ; d gS] nks oLr qvksa ds chp U; wur e
vyxko nwj h D; k gksxh] r kfd O; fDr dh vka[ k 500 nm
r j axnS/; Z fo; ksft r gS &
(1) 100 µm (2) 30 µm(3) 1 µm (4) 300 µm
Sol. 2 Dy 1.22d
9 2
2
500 10 25 102 0.25 10
y 30 m
27. , d bysDVªkWu , d gkbMªkst u t Sl s i j ek.kq@vk; u dh mÙksft rvoLFkk l s /kj kr y voLFkk r d l aØe.k dj r k gS %(1) bl dh xfr t Åt kZ c<+r h gS fdUrq fLFkfr t Åt kZ o dqyÅt kZ ?kVr h gSA(2) xfr t Åt kZ o dqy Åt kZ ?kVr h gS fdUr q fLFkfr t Åt kZc<+r h gSA(3) xfr t Åt kZ ?kVrh gS] fLFkfr t Åt kZ c<+rh gS fdUrq dqyÅt kZ l eku j gr h gSA(4) xfr t Åt kZ] fLFkfr t Åt kZ r Fkk dqy Åt kZ ?kVr h gSA
Sol. 1
ETotal = –13.6 eV 2
2
nZ
KE = |ETotal|
PE = 2 EtotalD; kasfd n ?kVr k gS] dqy Åt kZ ?kVr h gS] fLFkfr t Åt kZ ?kVr hgS r Fkk xfr t Åt kZ c<+r h gSA
28. l wph-I (ewyHkwr i z; ksx) dks l wph –II (bl ds l UnHkZ esa) l sfeykb, r Fkk nh xbZ l wph l s j [ kh fodYi pqfu, &List–I List–II
(A) Ýasad–gVZt + (i) i zdk' k dh d.kh; i zd frdk i z; ksx
(B) i zdk' k-fo| qr i z; ksx (ii) i j ek.kq ds for fj r Åt kZ Lr j
(C) Msfol u-t eZj i z; ksx (iii) bysDVªkWu dh r j ax i zd fr(iv) i j ek.kq dh l aj puk
(1) A i ; B iv ; C iii(2) A iv ; B iii ; C ii(3) A ii ; B i ; C iii(4) A ii ; B iv ; C iii
Sol. 3i zdk' k fo| qr i z; ksx i zdk' k dh d.kh; i zdfr l s l EcfU/kr gSA
29. 5 kHZ vkofÙk dk , d fl Xuy] 2 MHz vkofÙk dh okgd
r j ax i j vk; ke eksMwfyr gSA i fj .kkeh fl Xuy dh vkofÙk; k¡
gS@gSa %
(1) dsoy 2 MHz(2) 2000 kHz r Fkk 1995 kHz(3) 2005 kHz, 2000 kHz r Fkk 1995 kHz(4) 2005 kHz r Fkk 1995 kHz
JEE MAIN Examination(2015) (Code - A) (Page # 11)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. 3vkofÙk; k¡ gSFC, FC ± FS
30. , d LCR i fj i Fk esa voefUnr yksyd ds r qY; gSA , d LCRi fj i Fk esa l a/kkfj =k Q0 l s vkosf' kr gSA r c fp=kkuql kj L oR l s l a; ksft r gS :
R L
C; fn , d Nk=k L ds L1 o L2 (L1 > L2) ds nks fHkUu&fHkUuekukas ds fy, l e; t1 ds l kFk l a/kkfj =k i j vf/kdr e vkos' k(Q2
max) ds oxZ ds xzkQ vkj sf[ kr dj r k gS] r c fuEu eas l sdkSul k xzkQ l gh gS \ (vkj s[ k O; ofLFkr gS r Fkk i Sekuk ughafy; k gSA)
(1)
Q2Max
t
L1
L2 (2)
Q2Max
t
(For both L and L )1 2Q0
(3)
Q2Max
t
L1
L2 (4)
Q2Max
t
L1
L2
Sol. 1
KVL
R L
C
dI qIR L 0dt C
2
2
d q dq qL Rdt Cdt
voefUnr nksyu dh l ehdj .k ds l kFk r qyuk dj us i j
2
2
d y dyd kydtdt
vk; ke dh l ehdj .k bty Ae
t gk¡ Rb
2m 2L
Rt2L
max 0q q e
Rt2 2 Lmax 0q q e
l e; dky = RL
pwafd L1 > L2
1 < 2 vr % l gh mÙkj 3 gSAoSdfYi d gy %
Qmax dk eku ?kVr k gS] D; ksafd i zfr j ks/kd esa Åt kZ mRl ft Zrgksr h gSA D; ksasfd i zsjdRo dk eku c<+rk gS fuj kos' ku dh {kerki j ; k vkos' k c<+us i j A bl fy, Å"ek mRl t Zu l e; ?kVr kgSA vr %l gh xzkQ 3 gSA
JEE MAIN Examination(2015) (Code - A)(Page # 12)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
[CHEMISTRY]31. O; kol kf; d j sft au dk vkf.od l w=k C8H7SO3Na (vkf.od
Hkkj 206) gSa t ks fd i kuh dh l kf¶Vfuax ¼i kuh dh dBksj r k
gVkuk½ esa vk; uksa ds i fj or Zu ds fy, mi ; ksx esa vkr k gSaA
jsft ua ds } kj k Ca2+ vk; uksa dk vf/kdre eku D; k gksxk t c
j sft ua dks eksy@xzke esa cr k; k x; k gSa\
(1) 1031
(2) 2061
(3) 3092
(4) 4121
Sol. (4)2C8H7SO3Na + Ca7+ (C8H7SO3)2Ca2 eksy 1 eksy2 × 206 xzke Ca2+ dk 1 eksy ysr k gSA
1 xzke ysxk = 4121
Ca2+ ds eksy
32. dk; dsfUnzr ?kuh; t kyd esa l ksfM; e /kkrq fØLVyhdr gksrhgSa ft l dh , dd l sy dh yEckbZ 4.29 Å gSaA l ksfM; ei j ek.kq dh f=kT; k yxHkx gSa%(1) 1.86 Å (2) 3.22 Å(3) 5.72 Å (4) 0.93 Å
Sol. (1)
a3 = 4r
r =4
29.4732.1 = 1.86Å
33. gkbMªkst u dh l EHkkfor mÙksft r voLFkk dh Åt kZ fuEu esa l sdkSul h gSa\(1) +13.6 eV (2) –6.8 eV(3) –3.4 eV (4) +6.8 eV
Sol. (3)
2
2
nz6.13– gkbMªkst u ds fy, : z = 1
2n6.13–
l EHko gSA –13.6, –3.4, –1.5 etc.34. var j vkf.od vkd"kZ.k] i zfrd"kZ.k t ks fd O; qRde ?ku dh nwj h
i j fuHkZj dj r k gSa v.kqvksa ds e/; gSa&
(1) vk; u&vk; u cy
(2) vk; u&f} /kqzo vkd"kZ.k
(3) yanu cy
(4) gkbMªkst u ca/kSol. (4)
gkbMªkst u cU/k , d f} /kzqo & f} /kzqo i zHkko gSA
35. fuEufyf[ kr vfHkfØ; k 298 K i j gksr h gSa\2NO(g) + O2(g) 2NO2(g)NO(g) ds fuekZ.k dh ekud eqDr Åt kZ 86.6 kJ/mol gS]298 K i j A NO2(g) ds fuekZ.k dh ekud eqDr Åt kZ298 K i j D; k gksxh\ (Kp = 1.6 × 1012)(1) R(298 ln (1.6 ×1012) – 86600(2) 86600 + R(298) ln (1.6 × 1012)
(3) 86600 – )298(R)106.1ln( 12
(4) 0.5[2×86,600–R(298 ln (1.6 × 1012)]Sol. (4)
2106.1n298R–
12
= Gºr = 0NO
0NO G2–G2
2
0NO2
G =86.6×103 – 2
106.1nK298 12
36. , l hVksu dk ok"i nkc 20ºC i j 185 Vksj gSaA t c vok"i ' khyi nkFkZ dk 1.2 g, , fl Vksu ds 100 g esa ?kqyr k gS] 20ºCi j A bl dk ok"i nkc183 Vksj gSaA i nkFkZ dk eksyj nzO; eku(g mol–1) gSa%(1) 32 (2) 64(3) 128 (4) 488
Sol. (2)Pº = 185
PP–P0
= Nn
183183185
= 58/100M/2.1
M = 64
37. 2A B + C vfHkfØ; k ds fy, ekud fxCt Åt kZ
esa i fj or Zu 300K i j 2494.2J gSaA fn; s x; s l e; i j
vfHkfØ; k feJ.k dk l a?kVu [A] = 21
, [B] = 2 r Fkk [C]
= 21
gSaA v fHkfØ; k fd l fn' kk esa t k; sxhA
[R = 8.314 J/K/mol, e = 2.718](1) vxz fn' kk D; ksafd Q > KC
(2) i ' p fn' kk D; ksafd Q > KC
(3) vxz fn' kk D; ksafd Q < KC
(4) i ' p fn' kk D; ksafd Q < KC
JEE MAIN Examination(2015) (Code - A) (Page # 13)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. (2)Gº i j 300K = 2494.2 J
2A B + C
Gº = –RT n K–2494.2 = –8.314 × 300 n KK = 10
Q = 2]A[]C][B[ = 4
21
212
2
.
Q > KC reverse direction.38. CuSO4 ds foy; u esa 2 QSj kMs fo| qr xqt kjh xbZ gSaA dSFkksM
i j t ek gksus okys Cu dk Hkkj gksxk&(Cu dk i j ek.kq nzO; eku = 63.5 amu)(1) 0 g (2) 63.5 g(3) 2g (4) 127g
Sol. (2)Cu2+ + Ze Cu2 eksy bDdBk dj r k gS = 1 eksy Cu2F 2 eksy 1 eksy Cu 63.5 xzke
39. mPpr e Øe (>3) dh vfHkfØ; k nqyZHk gSa] dh ot g l s%(1) l Hkh vfHkfØ; k ; qDr Li h' kht dh , d l kFk VDdj ksa dhde l EHkkouk(2) , UVªksi h esa c<+ksr j h vkSj l fØ; .k Åt kZ t Sl s fd T; knkv.kq l fEefyr gSa(3) i zR; kLFk VDdj ksa ds dkj .k l kE; vfHkdeZd dh vksjfoLFkkfi r gksrk gSa(4) VDdj i j l fØ; Li h' kht dh gkfu
Sol. (1)vk.kfodr k vkSj dksfV > 3 l EHko ugh gSA D; ksfd l HkhvfHkfØ; k djus okys vo; oks dk , d l kFk Vdjko dh l EHkkoukde gSA
40. i k=k esa l fØ; pkj dksy dk 3 g, 50 mL ds , l hfVd vEyfoy; u (0.06N) esa feyk; k x; kA , d ?kaVs ckn bl dksNfu=k fd; k x; k vkSj Nfu=k dk l keF; Z 0.042N i k; k Xk; kA, fl fVd vEy dh vo' kksf"kr ek=kk ¼pkj dksy dk i zfr xzke½ gSa&(1) 18 mg (2) 36 mg(3) 42 mg (4) 54 mg
Sol. (1)CH3COOH (0.06M)50 mlfeyh eksy = 50 × 0.06 = 3cps gq, feyh eksy = 50 × 0.042 = 2.1vo' kksf"kr feyh eksy = 0.9
vo' kksf"kr Hkkj = 3
60109.0 3– × 103
= 354
= 18 feyh xzke
41. N3–, O2– vkSj F– (Å esa) dh vk; fud f=kT; k Øe' k% gSa&(1) 1.36, 1.40 vkSj 1.71(2) 1.36, 1.71 vkSj 1.40(3) 1.71, 1.40 vkSj 1.36(4) 1.71, 1.36 vkSj 1.40
Sol. (3)l Hkh vo; oks esa bysDVªksuksa dh l a[ ; k l eku gSA ; fn i zksVksuksadh l a[ ; k T; knk gksxh r ks vkdkj NksVk gksxkA
42. Al ds fu"d"kZ.k ds fy, gkWy&gsj ksYV i zØe esa] fuEu esa l sdkSul k dFku vl R; gSa\(1) CO vkSj CO2 bl i zØe esa cur s gSa(2) Al2O3, CaF2 ds l kFk fefJr gksr s gS t ks fd feJ.k dsxyukad fcUnq dks de dj r k gSa vkSj pkydr k ykr k gSa(3) Al3+ dSFkksM i j vi pf; r gksr k gSa] Al cukus ds fy,(4) Na3AlF6 fo| qr vi ?kVuh dh r j g gS
Sol. (4)
43. H2O2 ds l anHkZ esa fuEufyf[ kr dFku l s] vl R; dFku dkspqfu; s\(1) ; g dsoy vkWDl hdkj d , t saV dh r j g O; ogkj djl dr k gS(2) ; g i zdk' k esa , Dl i kst j i j fo?kfVr gksr k gS(3) bl s IykfLVd esa j [ kuk i M+r k gS vkSj oSDl ykbu Xyklcksr y va/ksj s esa(4) bl s /kqy l s nwj j [ kk t kr k gS
Sol. (1); g vkWDl hdkj d o vi pk; d nksuks dh r j g dke dj r k gSA
44. fuEu esa l s fdl esa] {kkj h; enk /kkr q l yQsV dh gkbMªs' kuÅt kZ vf/kdr e gSa ml dh t kyd Åt kZ l s\(1) CaSO4 (2) BeSO4(3) BaSO4 (4) SrSO4
Sol. (2)dsoy BeSO4 ?kqyu' khy l YQsV gSA D; ks fd bl dh t kydÅt kZ bl dh gkbMªs' ku Åt kZ l s de gksr h gSA buds vykokckdh l c vo{ksi h; gksr s gSA
45. fuEu esa l s dkSul k l cl s T; knk vfØ; gSa\(1) Cl2 (2) Br2(3) I2 (4) ICl
Sol. (4)bl ds fy, f} /kqzo vk/kq.kZ ' kqU; ugh gS] vr% ; g /kqzoh; gSA ckfdl c v/kqzoh; gSA r Fkk = 0.
JEE MAIN Examination(2015) (Code - A)(Page # 14)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
46. mRi szj d dk feyku dj ks] l gh i zØe ds l kFk%mRi szj d i zØe
(A) TiCl3 (i) osdj i zØe
(B) PdCl2 (ii) ft xy j &ukVk
cgqydhdj .k
(C) CuCl2 (iii) l Ei dZ i zØe
(D) V2O5 (iv) Mhdku i zØe(1) (A) - (iii), (B) - (ii), (C) - (iv), (D) (i)(2) (A) - (ii), (B) - (i), (C) - (iv), (D) (iii)(3) (A) - (ii), (B) - (iii), (C) - (iv), (D) (i)(4) (A) - (iii), (B) - (i), (C) - (ii), (D) (iv)
Sol. (2)
47. fuEu esa l s fdl dk DoFkukad fcUnq mPp gSa\(1) He (2) Ne(3) Kr (4) Xe
Sol. (4)i j ek.kq Hkkj ft r uk T; knk gksxk DoFkukad fcUnq mr uk ghT; knk gksxkA
48. T; kfefr l eko; fo; ksa dh l a[ ; k t ks fd oxkZdkj l er yh;[Pt(Cl)(py) (NH3) (NH2OH)]+ ds fy, fo| eku gSa(py = i sj hfMu) :(1) 2 (2) 3(3) 4 (4) 6
Sol (2)dsp2 Mabcdbl fy, T; kfer h; l eko; oh; ks dh l a[ ; k = 3
49. KMnO4 ds j ax dk dkj .k gSa&(1) M L vkos' k LFkkukUr j .k l aØe.k(2) d - d l aØe.k(3) L M vkos' k LFkkukUr j .k l aØe.k(4) * l aØe.k
Sol. (3)vkos' k dk LFkkukUr j .k yhxsUMl ~ l s /kkr q i j gksr k gSA bl fy,KMnO4 dk j ax cSxuh gksr k gSA
50. dFku : ok; qe.My esa ukbVªkst u vkSj vkWDl ht u eq[ ; ?kVdgS ysfdu ; s vfHkfØ; k dj ds ukbVªkst u dk vkWDl kbM ughacukr s gSaAdkj .k: ukbVªkst u vkSj vkWDl ht u ds e/; vfHkfØ; k ds fy,mPp r ki dh t : j r gSaA(1) nksuksa dFku vkSj dkj .k l gh gSa ysfdu dkj .k] dFku dsfy, l gh O; k[ ; k ugha dj r k gSaA(2) nksuksa dFku vkSj dkj .k l gh ugha gSaA(3) dFku l gh ugha gSa ysfdu dkj .k l gh gSaA(4) nksuksa dFku vkSj dkj .k l gh gSa vkSj dkj .k] dFku ds fy,l gh O; k[ ; k dj r k gSaA
Sol. (1)
51. gSykst uksa dh x.kuk dh D; wfj ; l fof/k esa] dkcZfud ; kSfxd ds250 mg, AgBr ds 141 mg nsr k gSaA ; kSfxd esa czksfeudk i zfr ' kr gSa% (Ag dk i j ek.kq nzO; eku = 108, Br = 80)(1) 24 (2) 36(3) 48 (4) 60
Sol. (1) Br ds eksy = 1 × AgBr ds eksy
= 1 × 188
10141 3–
Br dk Hkkj = 80188
10141 3–
Br dh % = 10010250
80188
101413–
3–
= 24%
52. fuEu esa l s dkSul k ; kSfxd T; kfefr ; l eko; ork esa fo| ekugSa\(1) 1-Qsfuy-2-C; wfVu(2) 3-Qsfuy-1-C; wfVu(3) 2-Qsfuy-1-C; wfVu(4) 1, 1-MkbZ Qsfuy-1-i zksi su
Sol. (1)H3C – HC = CH–CH2 – Phnksuks f} cU/k dkcZu i j ek.kq vyx&vyx f} i zfr LFkkfi r gSA
53. dkSul k ; kSfxd vkst ksfudj .k dj us i j 5-dhVks-2-esfFkygsDl susy nsxk\
(1)
CH3
CH3 (2)
CH3
CH3
(3)
CH3
CH3
(4) H3C
CH3
Sol. (2)
CH3 CH3
CH3 CH3
vkst ksuhdj .k OCHO
4
3 2
1
5
6CH3
CH3
JEE MAIN Examination(2015) (Code - A) (Page # 15)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
54. , Ydkby ¶yksj kbM dk l a' ys"k.k l cl s vPNk gksr k gSa] fdl ds}kj k&(1) eqDr ewyd ¶yksj huhdj .k(2) l sUMes; j vfHkfØ; k(3) ÝsUdyfLaVu vfHkfØ; k(4) LokZVl vfHkfØ; k
Sol. (4)R–Cl or
R – BrR–F + AgCl/AgBrAgF/dmF
Swart reaction
55. fuEufyf[ kr vfHkfØ; k ds Øe esa
VkWyqbZu A B
C
KMnO4 SOCl2
H /Pd2 BaSO4
mRi kn C gSa%(1) C6H5COOH (2) C6H5CH3(3) C6H5CH2OH (4) C6H5CHO
Sol. (4)COOH
(A)
(B)(C)
CHO
CH3
COCl
SOCl2
KMnO4
H /Pd2
BaSO4
56. vfHkfØ; k esaNH2
CH3
NaNO /HCl2
CuCn/KCN
D
E + N2
0 - 5°C
mRi kn E gSa%
(1)
COOH
CH3
(2) H3C CH3
(3)
CN
CH3
(4)
CH3
Sol. (3)NH2 N Cl2 CN
CH3 CH3 CH3
NaNO /HCl2 CuCN/KCN0-5ºC + N2
(E)(D)57. fuEu esa l s dkSul k cgqyd i saV vkSj ysDoj ds fuEkkZ.k esa
mi ; ksxh gSa\
(1) csdsykbV (2) fXyi Vsy
(3) i ksyhi zksfi u (4) i ksyhfoukby Dyksj kbMSol. (2)
fXyIVy cgqyd i saUV o ysdj cukus esa mi ; ksx gksr k gSA
58. fuEu esa l s dkSul k foVkfeu i kuh esa ?kqyu' khy gSa\
(1) foVkfeu C (2) foVkfeu D
(3) foVkfeu E (4) foVkfeu KSol. (1)
dsoy foVkfeu B o C i kuh esa ?kqyu' khy gSA buds vykok
foVkfeu ol k ?kqyu' khy gksr s gSA
59. fuEu esa l s dkSul k ; kSfxd i zfr vEy ugha gSa?
(1) , sY; qfefu; e gkbMªksDl kbM
(2) l hesVhMhu
(3) QSuyt hu
(4) j sfuVhMhuSol. (3)
fQusyt kbu i zfr vEy ugh gSA
60. fuEufyf[ kr esa l s fdl ; kSfxd dk j ax i hyk ugha gSa\(1) Zn2[Fe(CN)6](2) K3[Co(NO2)6](3) (NH4)3 [As(Mo3O10)4](4) BaCrO4
Sol. (1)(1) (NH4)3 [As(Mo3O10)4] = i hyk
(2) BaCrO4 = i hyk
(3) Zn2[Fe(CN)6] = l Qsn
(4) K3[Co(NO2)6] = i hyk
JEE MAIN Examination(2014) (Code - A)(Page # 16)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
[MATHEMATICS]
61. ekuk A r Fkk B nks l eqPp; gS ft l esa Øe' k% pkj r Fkk nks
vo; o gS r c mi l eqPp; A × B dh l a[ ; k Kkr dhft ; s]
ft l esa çR; sd esa de l s de r hu vo; o gksA(1) 219 (2) 256(3) 275 (4) 510
Sol. 1n(A × B) = 8Total subsets = 28
8C0 + 8C1 + 8C2
= 37No. of Req. Subsets = 256 – 37 = 219.
62. , d l fEeJ l a[ ; k z dks eki kad , d okyh l fEeJ l a[ ; k dgk
t kr k gS ; fn |z| = 1 ekuk z1 r Fkk z2 bl çdkj l fEeJ
l a[ ; k, gSA fd 1 2
21
z –2z
2 – z z eki kad , d okyh l fEeJ l a[ ; k
r Fkk z2 eki kad , d okyh l fEeJ l a[ ; k ugha gSA r c fcUnq z1
fLFkr gksxkA
(1) f=kT; k 2 dk oÙk
(2) f=kT; k 2 dk oÙk
(3) x-v{k ds l ekukUr j l j y j s[ kk
(4) y-v{k ds l ekukUr j l j y j s[ kkSol. 3
1 2
1 2
z –2z1
2 – z z
(z1 – 2z2) 1 2 1 2 1 2z – 2z 2 – z z 2 – z z
221 1 2 2 1 2 1 2 1 2| z | –2z z – 2z z 4 z 4 – 2z z – 2z z
2 21 2z z
2 2 2 21 2 1 2z z – z – 4 z 4 0
2 21 2z – 4 z –1 0
|z1| = 2
63. ekuk r Fkk l ehdj .k x2 – 6x – 2 = 0 ds ewy gS ; fn
n 1 ds fy , an = n – n r c 10 8
9
a – 2a2a dk eku
cj kcj gksxkA(1) 6 (2) –6(3) 3 (4) –3
Sol. 3
x2 – 6x – 2 = 0 2 – 6 – 2 = 0
2–6– 2 = 0 .....(1)an = n – n n 1a40 – 298 = 10 – 10 – 28 +28
= 8 (a2 – 2) – b8 (b2 – 2)= 8(6)– 8(6) (using (1)= 69 – 69
= 6a9now
10 8 9
9 9
a 2a 6a3
2a 2a
64. ; fn A =
1 2 22 1 –2a 2 b
, d vkO; wg gS t ks fd l ehdj .k
AAT = 9I dks l r qa"V dj r k gS t gk¡ I , 3 × 3 Øe dk
bdkbZ vkO; wg gS r c Øfer ; qXe (a, b) cj kcj gksxkA(1) (2, – 1) (2) (– 2, 1)(3) (2, 1) (4) (– 2, – 1)
Sol. 4AAT = 9 I
1 2 2 1 2 a 9 0 02 1 –2 2 1 2 0 9 0a 2 b 2–2b 0 0 9
a + 4 + 2b = 0 a + 2b = – 4 ....(i)2a + 2 – 2b = 0 a – b = – 1 ....(ii)From i and ii3b = – 3 b = – 1a = – 2
65. ds l Hkh ekuks dk l eqPp; ft l ds fyl j s[ kh; l ehdj .kksa dk
fudk;
2x1 – 2x2 + x3 = x1
2x1 – 3x2 + 2x3 = x2
– x1 + 2x2 = x3
vl axkeh gy j [ kr k gS &
(1) fj Dr l eqPp; j [ kr k gSA
(2) , dy l eqPp; j [ kr k gSA(3) nks vo; o j [ kr k gSA
(4) nks l s vf/kd vo; o j [ kr k gSASol. 3
= (2 – ) (2 + 3– 4) + 2 (– 2 + 2) + 1 (4– 3 – ) = 0– 3 – 2 + 6 + 8 – 3 – – 8 = 0– 3 – 2 + 5 – 3 = 0
JEE MAIN Examination(2014) (Code - A) (Page # 17)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
3 + 2 – 5 + 3 = 0( – 1) (2 + 2 – 3) = 0( – 1) ( + 3) ( – 1) = 0
= 1, 1, –3
66. 6,000 l s cM+s i w.kkZdksa dh l a[ ; k t ks fd vadks 3, 5, 6, 7r Fkk 8 ds mi ; ksx l s cukbZ t k l dr h gSA t cfd vadksa dh
i quj kofr u gksA(1) 216 (2) 192(3) 120 (4) 72
Sol. 26/7/8– – – –3 × 4C3 × 3! = 72– – – – – = 120Total = 192
67. f} i n i zl kj 501–2 x es x ds i w.kkZadh; ?kkr ks ds xq.kkdks
dk ; ksx gksxk &
(1) 501 3 12
(2) 501 32
(3) 501 3 –12
(4) 501 2 12
Sol. 1for sum of integral power of xput x = 1 in
50 501 2 x 1 2 x
2
503 12
.
68. ; fn m nks fofHkUu okLr fod l a[ ; kvksa l r Fkk n(l, n > 1)dk l ekUr j ek è; gS r Fkk G1, G2 r Fkk G3, l r Fkk n, ds
eè; r hu xq.kksÙkj ekè; gSA r c 4 4 41 2 3G 2G G cj kcj
gksxkA(1) 4 l2mn (2) 4 lm2n(3) 4 lmn2 (4) 4 l2m2n2
Sol. 2
m = n
2
2m = + r4
G1 G2 G3 n r r2 r3 r4 = n4r4 + 24r8 + 4r12
4r4(1 + 2r4 + r8)4r4 (1 + r4)2
4r4 22m
n · 3 2
24m
4m2n
69. Js<+h 311
+ 3 31 21 3
+ 3 3 31 2 31 3 5
+ ..... 9 ds
çFke 9 i nksa dks ; ksx gksxkA(1) 71 (2) 96(3) 142 (4) 192
Sol. 2
Tn = 3n
(2n 1)
=
2 2
2n (n 1)
4 n
Tn = 14 2n 2 n 1
= 14
n(n 1)(2n 1) 2n(n 1) n
6 2
= 14
9 10 19 90 96
= 14
{285 + 99} = 96
70. x 0Lim
(1– cos2x)(3 cos x)x tan4x
cj kcj gSA
(1) 4 (2) 3
(3) 2 (4) 12
Sol. 3
x 0Lim
(1 cos2x) (3 cos x)x tan4x
x 0Lim
(1 cos2x)· (3 cosx)tan4xx .4x
4x
JEE MAIN Examination(2014) (Code - A)(Page # 18)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
x 0Lim
2(1 cos2x) (3 cosx).
tan4x(2x)4x
12
·(3 + 1) = 2
71. ; fn Qyu g(x) = k x 1, 0 x 3mx 2, 3 x 5
vodyuh;
gS] r c k + m dk eku gksxkA
(1) 2 (2) 165
(3) 103 (4) 4
Sol. 1
g(x) = k x 1 x 0,3
mx 2 x 3,5
g(x) diff g(x) continuous g(3–) = g (3+)
k 4 = 3m + 2 2k = 3m + 2 ......(1)
Againg’(3+) = g’ (3–)
m = x 3
k k42 x 1
4m = k.....(2)from (1) & (2)2k = 3m + 2 8m = 3m + 2
5 m = 2
m = 25
& k = 4m = 85
k + m = 105
= 2
72. oØ x2 + 2xy – 3y2 = 0 i j fLFkr fcUnq (1, 1) i j
vfHkyEc
(1) oØ dks i qu% ugha feyr k gSA
(2) oØ dks i qu% f} r h; pr qZFkka' k es feyr k gSA
(3) oØ dks i qu% r r h; pr qZFkka' k esa feyr k gSA
(4) oØ dks i qu% pr qZFk pr qFkkZa' k esa feyr k gSASol. 4
x2 + 2xy – 3y2 = 0diff. w.r.t. x2x + 2x (y') + 2y – 6yy' = 02 + 2y' + 2 – 6y' = 0
4y' = 4y' = 1
slope of normal = – 1So equation becomesy – 1 = –1 (x – 1)x + y = 2
solving it with curvex2 + 2xy – 3y2 = 0x2 + 2x(2 – x) – 3(2 – x)2 = 0x2 + 4x – 2x2 – 3(x2 – 4x + 4) = 0–4x2 + 16x – 12 = 0x2 – 4x + 3 = 0(x – 1) (x – 3) = 0x = 1, 3
y = 1, –1thus second point of intersection is (3, –1)is in 4th qud.
73. ekuk f(x) pkj ?kkr dk , d cgqi n gS] ft l ds pj e eku
x = 1 r Fkk x = 2 i j gSA ; fn x 0Lim 2
f(x)1
x
= 3,
r c f(2) cj kcj gksxkA(1) –8 (2) –4(3) 0 (4) 4
Sol. 3f(x) =
x 0Lim 2
f(x)1x
= 3
f(x) must not contain degree 0 °ree 1 term f(x) = ax4 + bx3 + cx2
now f'(x) = 4ax3 + 3bx2 + 2cxf'(1) = 4a + 3b + 2c = 0 ......(1)f'(2) = 32a + 12b + 4c = 0 ......(2)
and x 0Lim 2
f(x)1x
= 1 + c = 3 ......(3)
JEE MAIN Examination(2014) (Code - A) (Page # 19)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
c = 2 12b 24
b 21a2
(1) 4a + 3b = – 4(2) 32a + 12b = – 8(1) 32a + 24b = – 32
74. l ekdyu 2 4 3/4dx
x (x 1) cj kcj gSA
(1)
1/44
4x 1 c
x
(2) 1/44x 1 c
(3) – 1/44x 1 c (4) –
1/44
4x 1 c
x
Sol. 4
2 4 3 /4dx
x (x 1)
= 5 4 3 /4dx
x (1 x ) = 5
4 3 / 4x dx
(1 x )
...(1)
put 1 + x–4 = T4
– 4x–5 d x= 4T3 dT (1) become
– 3
3T dTT = – T + C
= – (1 + x–4)1/4 + C
= –
1/44
41 x C
x
75. l ekdyu
4 2
2 22
logxlogx log(36 –12x x ) dx cj kcj
gSA(1) 2 (2) 4(3) 1 (4) 6
Sol. 3
4 2
222
log xI dxlogx log x 6
.....(1)
using b b
a a
f x dx = f a b x dx
I =
24
2 22
log 6-xdx
log 6 x logx ....(2)
(1) + (2) gives
2I = 4
2
1dx 2I = 1
76. nh?kZoÙk 2x9
+ 2y5
= 1 dh ukfHk; t hok ds vafr e fcUnqvksa
i j Li ' kZ j s[ kk } kj k cus pr qHkZqt dk {ks=kQy ¼oxZ bZdkbZ esa ½gksxkA
(1) 274
(2) 18
(3) 272
(4) 27
Sol. 4
P
P1
35
,2
2 2x y 19 5
a = 3 b = 5
e2 = 1 –
2
2ba
= 1 – 5 49 9
e = 23
now the quadrilateral formed will be a rhombus
with area = 22a
e
= 2.9 32
= 27
77. ekuk y(x) vody l ehdj .k (x log x) dydx + y = 2x
log x, (x 1) dk gy gS] r c y(e) cj kcj gksxkA(1) e (2) 0(3) 2 (4) 2e
Sol. 3
JEE MAIN Examination(2014) (Code - A)(Page # 20)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
dydx
+ 1
x log x . y = 2
I.F. = 1
log log xx logxe e log x
y· log x = 2·log x dx
y log x = 2 (x log x – x ) + cx = 1 c = 2x = e y = 2(e – e) + 2 = 2
78. mu fcUnqvksa dh l a[ ; k ft uds nksuksa funsZ' kakd i w.kkZad gSA t ks
' kh"kZ (0, 0), (0, 41) r Fkk (41, 0) l s cus f=kHkqt ds
vUnj fLFkr gS] gksxhA(1) 901 (2) 861(3) 820 (4) 780
Sol. 4
x + y < 411 to eachx + y < 39 x + y 38 x + y + z = 38
38 + 3 –1C3–1 = 40C2 = 40 39 780
2
.
79. fcUnq (2, 3) ds çfr fcEc dk j s[ kk (2x – 3y + 4) + k (x– 2y + 3) = 0, k R esa fcUnqi Fk gksxkA
(1) x-v{k ds l ekUr j l j y j s[ kk
(2) y-v{k ds l ekUr j l j y j s[ kk
(3) 2 f=kT; k dk oÙk
(4) 3 f=kT; k dk oÙk
Sol. 3
PR = RQ(x – 1)2 + (y – 2)2 = (2 – 1)2 + (3 – 2)2
(x – 1)2 + (y – 2)2 = 2
80. oÙkksa x2 + y2 – 4x – 6y – 12 = 0 r Fkk x2 + y2 +6x + 18y + 26 = 0, i j mHk; fu"B Li ' kZ j s[ kkvksa dh
l a[ ; k gSA(1) 1 (2) 2(3) 3 (4) 4
Sol. 3
x2 + y2 – 4x – 6y – 12 = 0
C1 (2, 3) r1 = 2 22 3 12 = 5
x2 + y2 + 6x + 18y + 26 = 0
C2 (–3, –9), r2 = 2 23 9 26 = 8
C1C2 = (52 + 122) = 13
81. {(x, y) : y2 2x r Fkk y 4x – 1} ds } kj k i fj Hkkf"kr
{ks=k dk {ks=kQy ¼oxZ bdkbZ esa½ gksxkA
(1) 732 (2)
564
(3) 1564 (4)
932
Sol. 4
y2 = 2x
2y y 12 4
2y2 – y – 1 = 02y2 – 2y + y – 1 = 0(2y + 1) (y – 1)
JEE MAIN Examination(2014) (Code - A) (Page # 21)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
21–12
y 1 yA – d4 2
1213
1–21–
2
y y y2A –4 6
1 12 3
1 1– –2 2
y 2y yA –8 6
1 –13 1 14A – –8 8 6 48
3 3 8 1 12 3 9A – –
8 32 48 32 48
15 9 3 5 3A – –
32 48 16 2 3
= 3 15 – 6 3 9 9
16 6 16 6 32
82. ekuk i j oy; x2 = 8y i j O ' kh"kZ r Fkk Q dksbZ fcUnq gS]
; fn fcUnq P j s[ kk[ k.M OQ dks 1 : 3 esa vUr % foHkkft rdj r k gSA r c P dk fcUnqi Fk gksxkA(1) x2 = y (2) y2 = x(3) y2 = 2x (4) x2 = 2y
Sol. 4
Let P : (h, k)
1. .0h 4h4
1. 3.0k 4k4
(,) on Parabola 2 = 8 (4h2) = 8.4 k
16h2 = 32kx2 = 2y
83. fcUnq (1, 0, 2) dh j s[ kk x – 2
3 = y 1
4
= z – 212
r Fkk
l er y x – y + z = 16 ds çfr PNsnh fcUnq l s nwj h Kkr
dhft ; sA
(1) 2 14 (2) 8
(3) 3 21 (4) 13
Sol. 4P(3 + 2, 4 – 1, 12 + 2)3 + 2 – 4 + 1 + 12 + 2 = 1611 = 11 = 1Point of intersection (5, 3, 14)
Distance = 2 2 24 3 12
= 169= 13
84. ml l er y dh l ehdj .k Kkr dhft ; s ft l esa j s[ kk
2x – 5y + z = 3; x + y + 4z = 5 gS r Fkk l er y
x + 3y + 6z = 1 ds l ekukUr j gSA(1) 2x + 6y + 12z = 13(2) x + 3y + 6z = – 7(3) x + 3y + 6z = 7(4) 2x + 6y + 12z = –13
Sol. 32x – 5y + z – 3 + (x + y + 4z – 5) = 0x( + 2) + y( – 5) + z(4 + 1) – 3 – 5 = 0
2 5 4 11 3 6
3 + 6 = – 5 6 – 30 = 12 + 32 = – 11 6 = – 33 = – 11/2 = – 11/24x – 10y + 2z – 6–11x – 11y – 44z + 55 = 0– 7x – 21y – 42z + 49 = 0x + 3y + 6z – 7 = 0
JEE MAIN Examination(2014) (Code - A)(Page # 22)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
85. ekuk a,b
r Fkk c
r hu v' kqU; l fn' k bl i zdkj gS fd mues
l s dksbZ nks l j s[ kk, sa r Fkk (a b
) × c
=13 b
c
a.
; fn l fn' k b r Fkk c
ds e/; dk dks.k gS] r c sin dk
eku gksxk &
(1) 2 2
3(2)
– 23
(3) 23 (4)
–2 33
Sol. 11(a b) c bc a3
1(a.c) b – (b.c) a bc a 0.b3
1–b.c bc3
, a.c 0
– bc cos = 1 bc3
cos = 1–3
sin = 2 23
86. ; fn 12 , d l eku xsans 3 , d l eku fMCcks esa Mkyh t kr hgS] r c i zkf; dr k Kkr dhft , t cfd bu fMCCkksa esa l s , d esaBhd 3 xsans gksA
(1) 11
55 23 3
(2) 10
2553
(3) 12
12203
(4) 11
1223
Sol. 1
Success (p) = 13
Failure (q) = 23
Acc. to binomial distribution, we have to find,
P(x = 3) = 3C12 .
313
· 9
23
= 553
1123
.
87. 16 çs{k.kks ds vk¡dM+ksa ds l eqP; dk ekè; 6 gS ; fn buesa l s, d çs{k.k eku 16 dks gVk fn; k t kr k gS r Fkk u; s r hu 3,4 r Fkk 5 eku ds çs{k.kksa dks vk¡dM+ksa esa t ksM+k t kr k gS r ci fj .kkeh vk¡dM+ks dk ekè; gksxkA(1) 16.8 (2) 16.0(3) 15.8 (4) 14.0
Sol. 4
1 2 3 15a a a ...... a 1616
16
......(1)
1 2 3 15a a a ...... a (3 4 5)??
18
......(2)(1) a1 + a2 + a3 +.....+ a15 = (16)2 – 16
now
(2) 2(16) 16 12
18
= 256 4
18
= 25218
= 14
mean = 14
88. ; fn fdl h Vkoj ds f' k"kZ l s r hu l j s[ kk, sa fcUnq A, B r Fkk CVkoj ds i kn dh vksj t kus okyh j s[ kk l s mUu; u dks.k Øe' k%30º, 45º r Fkk 60º cukr s gS r c AB : BC dk vuqi krgksxk &
(1) 3 :1 (2) 3 : 2(3) 1: 3 (4) 2 : 3
JEE MAIN Examination(2014) (Code - A) (Page # 23)
Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015
Sol. 1
AB h( 3 –1) 3 :11BC h(1– )3
89. ekuk tan–1y = tan–1x + tan–12
2x1 – x
,
t gk¡ |x| < 13 r c y dk , d eku gksxkA
(1) 3
23x – x1– 3x
(2) 3
23x x1 – 3x
(3) 3
23x – x1 3x
(4) 3
23x x1 3x
Sol. 1
tan–1 y = tan–1 x + tan–1 22x
1 x
|x| < 13
tan–1 22x
1 x = 2 tan–1 x
tan–1 y = tan–1 x + 2 tan–1 x= 3 tan–1 x
= tan–1 3
23x x1 3x
y = 3
23x x1 3x
90. ~ s v(~ r ^ s) dk udkj kRed cj kcj gSA(1) s ^ ~ r (2) s ^ (r ^ ~ s)(3) s v (r v ~ s) (4) s ^ r
Sol. 4~ S V (~ r ^ S)
S r ~ r ~ r ^ S ~ S ~ S V (~ r ^ S) ~ (~ SV(~ r ^ S))T T F F F F TT F T T F T FF T F F T T FF F T F T T F