ALLEN JEE-MAIN SAMPLE PAPER # 01 TARGET - 2014

Transcription

ALLEN
TM
CAREER INSTITUTE
Path to Success KOTA (RAJASTHAN)
ALLEN JEE-MAIN SAMPLE PAPER # 01
TARGET - 2014
egRoiw . kZ lw p uk,¡
IMPORTANT INSTRUCTIONS
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iq fLrdk dks rc rd u [kksysa tc rd dgk u tk,A
ijh{kk iqfLrdk ds bl i`"B ij vko';d fooj.k uhys@dkys ckWy ikbaV isu
ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gaSA
ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkj
i= ij dgha vkSj u fy[ksaA
ijh{kk dh vof/k 3 ?ka V s gSA
bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 360 gSaA
1.
Immediately fill in the form number on this page of the Test Booklet
with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.
1.
2.
The candidates should not write their Form Number anywhere else
(except in the specified space) on the Test Booklet/Answer Sheet.
2.
3.
The test is of 3 hours duration.
3.
4.
The Test Booklet consists of 90 questions. The maximum marks are
360.
4.
5.
There are three parts in the question paper A,B,C consisting of
Mathematics, Physics and Chemistry having 30 questions in each
part of equal weightage. Each question is allotted 4 (four) marks for
correct response.
5.
bl ijh{kk iqfLrdk es a rhu Hkkx A, B, C gSa] ftlds izR;sd Hkkx esa
xf.kr] HkkSfrd foKku ,oa jlk;u foKku ds 30 iz'u gaS vkSj lHkh iz'uksa ds vad
leku gASa izR;sd iz'u ds lgh mÙkj ds fy, 4 (pkj)vad fuèkkZfjr fd;s x;s gAaS
6.
One Fourth mark will be deducted for indicated incorrect response
of each question. No deduction from the total score will be made
if no response is indicated for an item in the Answer Sheet.
6.
7.
Use Blue/Black Ball Point Pen only for writting particulars/marking
responses on Side–1 and Side–2 of the Answer Sheet.
Use of pencil is strictly prohibited.
7.
8.
No candidate is allowed to carry any textual material, printed or written,
8.
izR;sd xyr mÙkj ds fy, ml iz'u ds dqy vad dk ,d pkSF kkbZ vad dkVk
tk;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls
½.kkRed vadu ugha gksxkA
mÙkj i= ds i` " B&1 ,oa i` " B&2 ij okafNr fooj.k ,oa mÙkj vafdr djus gsrq
dsoy uhys@ dkys ckWy ikba V isu dk gh iz;ksx djsaA
isf Uly dk iz ;ksx fcYdqy oftZr gSA
ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkh
izdkj dh ikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa] istj]
eksckby Qksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU;
izdkj dh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA
bits of papers, pager, mobile phone any electronic device etc, except
the Identity Card inside the examination hall/room.
9.
jQ dk;Z ijh{kk iqfLrdk esa dsoy fu/kkZfjr txg ij gh dhft;sA
10. On completion of the test, the candidate must hand over the Answer
Sheet to the invigilator on duty in the Room/Hall. However, the
candidate are allowed to take away this Test Booklet with them.
10.
11. Do not fold or make any stray marks on the Answer Sheet.
11.
ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{k fujh{kd
dks vo'; lkiSa nsAa ijh{kkFkhZ vius lkFk bl ijh{kk iq fLrdk dks ys tk
ldrs gaS A
mÙkj i= dks u eksMa+s ,oa u gh ml ij vU; fu'kku yxk,saA
9.
Rough work is to be done on the space provided for this purpose in
the Test Booklet only.
Corporate Office
ALLEN Career Institute,
“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005,
Trin : +91 - 744 - 2436001 Fax : +91-744-2435003,
E-Mail: [email protected] Website: www.allen.ac.in
2014
HAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS
BEWARE OF NEGATIVE MARKING
PART A - MATHEMATICS
2.
If ƒ(x) = tan–1x + ln(1 + x2) – x, then ƒ(x) is 1.
increasing in the interval (1) (0, ¥)
(2) (–1, 1)
(3) (0, 2)
(4) (–¥, 0) È (2, ¥)
If " a Î R one root of equation
2.
2
2
2
x + 2ax + b – a – 6 = 0 is less than 1 and one
root is greater than 1, then range of 'b' is(1) (–¥,–2)
(2) (–2,2)
(
(3) - 5, 5
3.
)
(1) (0, ¥)
(2) (–1, 1)
(3) (0, 2)
(4) (–¥, 0) È (2, ¥)
2
;fn " a Î R lehdj.k x + 2ax + b2 – a2 – 6= 0
dk ,d ewy 1 ls de rFkk ,d ewy 1 ls cM+k gks] rks
'b' dk ifjlj gksxk&
(1) (–¥,–2)
(2) (–2,2)
(
(3) - 5, 5
(4) (2,¥)
x2
Value of lim
is x ® 0 sin(2 p sec x)
(1)
4.
;fn ƒ(x) = tan–1x + ln(1 + x2) – x gks] rks ƒ(x)
fuEu vUrjky esa o/kZeku gksxk&
A
LL
EN
1.
1
2p
(2)
1
p
(3)
2
p
3.
(4)
4
p
x2
dk eku gksxk&
x ® 0 sin(2 p sec x)
1
2p
Coefficient of x 5 in the expansion of
10
5.
9
1ö æ
1ö
æ
ç x + ÷ ç x - ÷ is xø è
xø
è
(1) 9C3
(2) 9C4
(3) – 9C3 (4) –9C4
SPACE FOR ROUGH WORK /
ALLEN
(2)
1
p
(3)
2
p
(4)
4
p
oØ y = (x – 2)2 rFkk js[kkvksa y = x o y = 0 }kjk
ifjc¼ {ks=Qy gksxk&
(1)
5
6
10
5.
(4) (2,¥)
lim
(1)
Area bounded by curve y = (x – 2)2 and lines 4.
y = x & y = 0, is 5
6
2
(2)
(3)
(4) 1
(1)
6
5
3
)
(2)
6
5
(3)
2
3
(4) 1
9
1ö æ
1ö
æ
5
ç x + ÷ ç x - ÷ ds izlkj esa x dk xq.kkad gksxk xø è
xø
è
(1) 9C3
(2) 9C4
(3) – 9C3
(4) –9C4
jQ dk;Z ds fy;s txg
H-1/31
2014
7.
8.
9.
2
2
2
Sum of series 2 + 2.3 + 3.4 + ...... + 9.10
is (1) 2440 (2) 2600 (3) 2640 (4) 2500
Let ƒ : R ® R be a differentiable function such
that ƒ(1) = 3ƒ'(1) = 2.
Let g(x) = ƒ(–11x + 3ƒ2(x)), then g'(1) is equal
to (1) 9
(2) 11
(3) –6
(4) –2
Equation of line through origin in the plane
P : x + 2y = 2z which intersects the
perpendicular drawn from points (5, 1, –1) to
the plane P, is (1) x = –y = z
(2) x = –4y = 4z
(3) 4x = –y = 4z
(4) 4x = – 4y = z
There are 3 bags A, B & C. Bag A contains
1 Red & 2 Green balls, bag B contains 2 Red
& 1 Green balls and bag C contains only one
green ball. One ball is drawn from bag A &
put into bag B then one ball is drawn from B
& put into bag C & finally one ball is drawn
from bag C & put into bag A. When this
operation is completed, probability that bag A
contains 2 Red & 1 Green balls, is -
6.
7.
2
2
2
Js.kh 2 + 2.3 + 3.4 + .. + 9.102 dk ;ksxQy
gksxk&
(1) 2440 (2) 2600 (3) 2640 (4) 2500
ekuk ƒ : R ® R vodyuh; Qyu bl izdkj gS fd
ƒ(1) = 3ƒ'(1) = 2 gAS
ekuk g(x) = ƒ(–11x + 3ƒ2(x)) gks] rks g'(1) cjkcj
gksxk&
(1) 9
8.
(2) 11
(3) –6
(4) –2
lery P : x + 2y = 2z esa ewyfcUnq ls xqtjus okyh
js[kk dk lehdj.k] tks fcUnq (5, 1, –1) ls lery P
A
LL
EN
6.
2
(1)
1
4
(2)
1
2
(3)
1
3
(4)
ij [khpsa x;s yEc dks izfrPNsn djrh g]S gksxh&
(1) x = –y = z
(2) x = –4y = 4z
(3) 4x = –y = 4z
(4) 4x = – 4y = z
9.
1
6
H-2/31
3 FkSys A, B rFkk C gAS Fky
S s A esa 1 yky ,oa 2 gjh
xsnsa] Fky
S s B esa 2 yky ,oa 1 gjh xsan rFkk Fky
S s C esa
dsoy 1 gjh xsan gSA FkSys A ls 1 xsan fudky dj
mldks Fky
S s B esa j[krs g]S mlh le; Fky
S s B ls 1 xsan
fudky dj Fky
S s C esa j[krs gS rFkk vUr esa Fky
S s C ls 1
xsan fudky dj Fky
S s A esa j[krs gAS ;g izfØ;k iw.k± gksus
ij izkf;drk rkfd Fky
S s A esa 2 yky rFkk 1 gjh xsan gks]
gksxh -
(1)
1
4
(2)
1
2
(3)
1
3
(4)
1
6
jQ dk;Z ds fy;s txg
ALLEN
2014
10.
If mean deviation of sample 1, 2, 3, x about 3
is 1 (where x > 3), then possible value of
variance is 1
2
(2)
1
4
(1)
1
2
(2)
1
4
(3)
5
4
(4)
5
2
(3)
5
4
(4)
5
2
2p
e |sin x|
dx is equal to e|sin x| + e|cosx|
11.
0
(1) 0
(2) 2p
(3) p
(4) 1
r
r
If a and b are two diagonals of a quadrilateral
r
r
r r rr
such that | a - b | = a.b , | a | = 1 , | b | = 2 ,
12.
then area of quadrilateral is (1)
13.
ò
e |sin x|
dx cjkcj gksxk&
e|sin x| + e|cosx|
A
LL
EN
ò
0
12.
;fn 3 ds lkis{k izfrn'kZ 1, 2, 3, x (tgk¡ x > 3) dk
ek/; fopyu 1 gks ] rks iz l j.k dk la Hko eku
gksxk&
(1)
2p
11.
10.
1
2
(2) 3
(3) 1
dk {ks=Qy gksxk-
(4)
3
2
If a curve passing through (1,1) satisfies the 13.
differential equation x2dy = y2dx + xyd(xy),
then ln|xy| is equal to (1)
1
1
- 2
2
x
y
1 1
(3) y x
(2)
1 1
x y
1 1
(4) 2 - 2
y x
ALLEN
(1) 0
(2) 2p
(3) p
(4) 1
r
r
;fn a rFkk b prqHkqZt ds nks fod.kZ bl izdkj gS fd
r
r r rr r
| a - b | = a.b , | a | = 1 , | b | = 2 gks] rks prqHkqZt
(1)
1
2
(2)
3
(3) 1
(4)
3
2
;fn fcUnq (1,1) ls xqtjus okyk oØ vody lehdj.k
x2dy = y2dx + xyd(xy) dks larq"V djrk gS] rks
ln|xy| cjkcj gksxk&
(1)
1
1
- 2
2
x
y
(2)
1 1
x y
(3)
1 1
y x
(4)
1 1
y2 x2
jQ dk;Z ds fy;s txg
H-3/31
2014
14.
Equation of a tangent to the curve 16xy = 9 14.
passing through a point equidistent from
coordinate axes on the curve y + x3 = 0 in
second quadrant, is (1) x + 4y = 3
(2) 2x + y = –1
(3) 2x + 3y = 1
ìe ax
x£0
ƒ(x)
=
Let
. If x = 0 is point
í
î b tan x + c x > 0
of local maxima, then -
15.
(4) x + y = 0
ìe ax
x£0
ƒ(x)
=
ekuk
gSA ;fn x = 0
í
î b tan x + c x > 0
(2) a > 0, b ¹ 0, c £ 1
(3) a < 0, b ¹ 0, c ³ 1
(3) a < 0, b ¹ 0, c ³ 1
(4) a > 0, b ¹ 0, c ³ 1
(4) a > 0, b ¹ 0, c ³ 1
If ƒn(q) = cosnq + sin2nq " n Î N, then number
of solution of ƒ1(q) = ƒ2(q) in q Î [0, 2p], is (1) 1
18.
(3) 2x + 3y = 1
(1) a < 0, b ¹ 0, c £ 1
(2) a > 0, b ¹ 0, c £ 1
17.
(2) 2x + y = –1
LFkkuh; mfPp"B fcUnq gks] rks -
(1) a < 0, b ¹ 0, c £ 1
16.
(1) x + 4y = 3
A
LL
EN
15.
(4) x + y = 0
oØ 16xy = 9 dh Li'kZ js[kk dk lehdj.k tks f}rh;
prqFkk±'k esa oØ y + x3 = 0 ij fLFkr fcUnq tks funsZ'khZ
v{kksa ls leku nwjh ij fLFkr g]S ls xqtjrh g]S gksxh&
(2) 2
(3) 3
16.
(4) 4
;fn ƒn(q) = cosnq + sin2nq " n Î N gks] rks
vUrjky [0, 2p] esa ƒ1(q) = ƒ2(q) ds gyksa dh la[;k gksxh-
(1) 1
(2) 2
(3) 3
(4) 4
If normal at point (h, k) on parabola y2 = 8x 17.
meets parabola again at point (18, 12), then
sum of all possible value of h, is -
;fn ijoy; y2 = 8x ds fcUnq (h,k) ij vfHkyEc
ijoy; dks iqu% fcUnq (18, 12) ij feyrk gS] rks h ds
lHkh laHko ekuksa dk ;ksxQy gksxk&
(1) 8
(1) 8
(2) 12
(3) 10
(4) 6
(2) 12
(3) 10
(4) 6
If tangents drawn from point (1, 2) to ellipse 18.
x2 + 2y2 = 1 cuts x-axis at points A(x1, 0) &
B(x2, 0), then H.M. of x1 & x2 is -
;fn fcUnq (1, 2) ls nh?kZo`Ùk x2 + 2y2 = 1 ij [khaph xbZ
Li'kZ js[kk;sa x- v{k dks fcUnq A(x1, 0) rFkk B(x2, 0)
ij dkVrh gS] rks x1 rFkk x2 dk gjkRed ek/; gksxk&
(1) –9
(1) –9
(2) 9
(3) –7
(4) 7
H-4/31
(2) 9
(3) –7
(4) 7
jQ dk;Z ds fy;s txg
ALLEN
2014
20.
21.
Equation of directrices of hyperbola whose foci 19.
are (6, 8) & (10, 8) and eccentricity 2, are 7
9
(1) x = 9, x = 7
(2) x = - , x =
2
2
17
15
(3) x = , x =
(4) x = ± 16
2
2
Number of integral values of m for which 20.
y-coordinate of point of intersection of lines
y = mx + 1 & x + y = 2 is also an integer, is (1) 0
(2) 1
(3) 2
(4) 3
Let PA & PB are tangents to circle 21.
S1 º x2 + y2 – 2x – 2y + 1 = 0. If circumcircle
of DPAB is circle S2 and circle S1 bisects
circumference of S2, then diameter of circle S2
is (1) 1
22.
23.
ml vfrijoy ; dh fu;rk dk lehdj.k] ftldh
ukfHk;k¡ (6, 8) ,oa (10, 8) rFkk mRdsUnzrk 2 g]S gksxh&
(1) x = 9, x = 7
(3) x =
17
15
,x=
2
2
(2)
(3) 2
(4) 3
2
Number of different words that can be formed 22.
from all letters of word APPLICATION such
that two vowels never come together is (1) (45)7! (2) 8!
(3) 6!7! (4) (32)6!
1
Height of tower AB is
of tower CD and 23.
4
distance between their feet is 50 3 m. If angle
7
9
(2) x = - , x =
2
2
(4) x = ± 16
m ds iw.kk±d ekuksa dh la [;k ftlds fy;s js[kkvksa
y = mx + 1 rFkk x + y = 2 ds izfrPNsn fcUnq dk y-
funsZ'kkad Hkh ,d iw.kk±d g]S gksxh&
(1) 0
(2) 1
(3) 2
(4) 3
2
2
ekuk PA rFkk PB o`Ùk S1 º x + y –2x–2y+1 = 0
ds Li'kZ js[kk;sa gSA ;fn f=Hkqt PAB dk ifjo`Ùk] o`Ùk
S2 gks rFkk o`Ùk S1, S2 dh ifjf/k dks lef}Hkkftr djrk
gks] rks o`Ùk S2 dk O;kl gksxk&
A
LL
EN
19.
(1) 1
(2)
2
(3) 2
(4) 3
'kCn APPLICATION ds lHkh v{kjksa ls fufeZr fd;s
tk ldus okys fHkUu 'kCnksa dh la[;k rkfd nks Loj
dHkh Hkh lkFk uk gks] gksxh&
(1) (45)7! (2) 8!
(3) 6!7!
(4) (32)6!
,d ehukj AB dh Å¡pkbZ ehukj CD dh Å¡pkbZ dh
of elevation of top of tower CD is double of
angle of depreciation of its foot both from top
of tower AB, then height of tower CD is -
1
gS rFkk buds iknksa ds e/; nwjh 50 3 m gAS ;fn
4
ehukj AB ds 'kh"kZ ls ehukj CD ds 'kh"kZ dk mUu;u
dks.k ogh ls ehukj CD ds ikn ds voueu dks.k dk
nksxquk gks]rks ehukj CD dh Å¡pkbZ gksxh&
(1) 50 m
(1) 50 m
(2) 100 m (3) 150 m (4) 200 m
ALLEN
(2) 100 m (3) 150 m (4) 200 m
jQ dk;Z ds fy;s txg
H-5/31
2014
24.
Domain of the function
24.
ƒ(x) = sin -1 (cos(tan -1 x)) + cos-1 (sin(cot -1 x))
is (1) R
(2) [–1, 1]
ƒ(x) = sin -1 (cos(tan -1 x)) + cos-1 (sin(cot -1 x))
dk izkar gksxk&
(4) ( -¥, -1] È [1, ¥)
(3) [0, 1]
(2) [–1, 1]
(3) [0, 1]
(4) ( -¥, -1] È [1, ¥)
é 1 + a 2 + a 4 1 + ab + a 2 b 2 1 + ac + a 2 c 2 ù
ê
ú
25. ;fn A = ê1 + ab + a 2 b2 1 + b2 + b 4 1 + bc + b2 c 2 ú
ê1 + ac + a 2 c 2 1 + bc + b 2 c 2 1 + c 2 + c 4 ú
ë
û
rFkk det(A) = det(4I), tgk¡ I, 3 × 3 dk rRled
vkO;wg g]S rks (a – b)3 + (b – c)3 + (c – a)3 dk eku
gks ldrk g-S
and det(A) = det(4I), where I is 3 × 3 identity
matrix, then (a – b)3 + (b – c)3 + (c – a)3 can be
equal to -
26.
(1) R
A
LL
EN
é 1 + a2 + a4 1 + ab + a2 b2 1 + ac + a2c2 ù
ê
ú
25. If A = ê1 + ab + a2 b2 1 + b2 + b4 1 + bc + b2c2 ú
ê1 + ac + a2c2 1 + bc + b2c2 1 + c2 + c4 ú
ë
û
Qyu
(1) –24
(2) 6
(3) –6
(4) 12
26.
If system of equation
(1) –24
(2) 6
(3) –6
(4) 12
;fn lehdj.k fudk;
(tanq)x + (cotq)y + (8cos2q)z = 0
(tanq)x + (cotq)y + (8cos2q)z = 0
(cotq)x + (8cos2q)y + (tanq)z = 0
(cotq)x + (8cos2q)y + (tanq)z = 0
(8cos2q)x + (tanq)y + (cotq)z = 0
(8cos2q)x + (tanq)y + (cotq)z = 0
have non trivial solution, then sin4q is equal
to -
ds vfujFkZd gy gks] rks sin4q dk eku gksxk&
3
(1) 2
(3) -
1
2
(1) -
(2) –1
1
(4)
2
H-6/31
(3) -
3
2
1
2
(2) –1
(4)
1
2
jQ dk;Z ds fy;s txg
ALLEN
2014
27.
Dual of negation of the statement "He is 27.
gS-
laborious and intelligent" is(1) He is not laborious and not intelligent
(1) oks uk rks ifjJeh rFkk uk gh cqf}eku gAS
(2) He is laborious or not intelligent
(2) oks ifjJeh gS ;k cqf}eku ugha gAS
(3) He is not laborious or not intelligent
(3) og ifjJeh ugha gS ;k cqf}eku ugha gSA
(4) He is not laborious and not intelligent
(4) og ifjJeh ugh gS rFkk cqf}eku ugha gSA
r
Statement-1 : If aˆ + 2bˆ + 3cˆ = 0 , then
28.
r
dFku -1 : ;fn aˆ + 2bˆ + 3cˆ = 0 gks] rks | aˆ + bˆ |= 1
A
LL
EN
28.
dFku ^ôg ifjJeh rFkk cqf}eku g*S * dk nksgjk izfrokn
ˆ = 1.
| aˆ + b|
gksxkA
and
rFkk
Statement-2 : Sum of two unit vectors is always
a unit vector.
dFku -2 : nks bdkbZ lfn'kksa dk ;ksxQy lnSo ,d bdkbZ
lfn'k gksxkA
(1) Statement-1 is True, Statement-2 is True ;
(1)
Statement-2 is a correct explanation for
Statement-1.
dFku-I dh lgh O;k[;k gSA
(2)
(3) Statement-1 is True, Statement-2 is False.
dFku -I lR; g S _ dFku -II lR; g S_ dFku -II
dFku-I dh lgh O;k[;k ugha gSA
Statement-2 is NOT a correct explanation for
Statement-1.
dFku -I lR; g S _ dFku -II lR; g S _ dFku -II
(3)
dFku-I lR; gS] dFku-II vlR; gSA
(4)
dFku-I vlR; gS] dFku-II lR; gSA
(4) Statement-1 is False, Statement-2 is True.
ALLEN
jQ dk;Z ds fy;s txg
H-7/31
30.
Statement-1 : z is a complex number which 29.
satisfies |z – 1 – i| = 1. If z1 & z2 are two distinct
values of z for which |z – 2| + |z – 2i| is
minimum, then |z1 – z2| = 2
and
Statement-2 : If z = 2 or z = 2i, then |z – 2| + |z –
2i| is minimum.
Statement-1.
Statement-1.
Statement-1 : log103 + log106 < log10(81/4). 30.
and
Statement-2 : If ƒ(x) = log10x, then ƒ"(x) < 0
" x Î R+
Statement-1.
Statement-1.
dFku -1 : z ,d lfEeJ la[;k g]S tks |z – 1 – i| = 1 dks
larq"V djrh gAS ;fn z1 rFkk z2, z ds nks fHkUu eku gS
ftlds fy;s |z – 2| + |z – 2i| U;w u re g S ] rks
|z1 – z2| = 2 gksxkA
rFkk
dFku -2 : ;fn z = 2 ;k z = 2i gks] rks
|z – 2| + |z – 2i| U;wure gksxkA
(1)
A
LL
EN
29.
2014
H-8/31
(2)
(3)
dFku -1 : log103 + log106 < log10(81/4).
rFkk
dFku -2 : ;fn ƒ(x) = log10x gks] rks ƒ"(x) < 0
" x Î R+ gksxkA
(4)
(1)
(2)
(3)
(4)
jQ dk;Z ds fy;s txg
ALLEN
2014
PART B - PHYSICS
32.
The relative density of a metal may be found 31.
by hanging a block of the metal from a spring
balance and noting that in air the balance reads
(5 ± 0.05) N while in water it reads
(4 ± 0.05) N. The relative density would be
quoted as:
fdlh /kkfRod CykWd dks fLizax rqyk ls yVdkdj /kkrq
dk lkisf{kd ?kuRo Kkr fd;k tk ldrk gAS ekuk ok;q
esa fLizax rqyk dk ikB~;kad (5 ± 0.05) N o ty esa
(4 ± 0.05)N izkIr gksrk gS rks bldk lkisf{kd ?kuRo
gksxk %&
(1) (5 ± 0.05)
(2) 5 ± 11%
(1) (5 ± 0.05)
(2) 5 ± 11%
(3) (5 ± 0.10)
(4) 5 ± 6%
(3) (5 ± 0.10)
(4) 5 ± 6%
A
LL
EN
31.
An electromagnetic wave of wavelength l0 32.
(in vacuum) passes from P towards Q crossing
three different media of refractive index m, 2m
and 3m respectively as shown in figure. fP and
fQ be the phase of the wave at points P and Q.
Find the phase difference fQ – fP.
[Take : m=1]
P
µ
(3)
p
2
2.25l0
3l0
3.5l0
µ
3µ
2µ
(2)
p
4
(4) p
ALLEN
P
Q
2.25l0
(1) 0
fuokZr esa rjaxn/S ;Z l0 okyh ,d fo|qr pqEcdh; rjax
fp=kuqlkj Øe'k% m, 2m rFkk 3m viorZukadksa okys
rhu fofHkUu ek/;eksa dks ikj djrs gq, fcUnq P ls fcUnq
Q dh rjQ xeu djrh gAS fcUnqvksa P rFkk Q ij rjax
dh dyk fP rFkk fQ gS] dykUrj fQ – fP Kkr dhft;sA
[m=1 ysa]
Q
3µ
2µ
(1) 0
(3)
3l0
3.5l0
p
2
(2)
p
4
(4) p
jQ dk;Z ds fy;s txg
H-9/31
In a long cylindrical vessel made of perfectly 33.
conducting walls, an ideal mono-atomic gas is
confined with the help of a light piston, which can
slide inside the cylinder without friction. Number
of atoms of the gas in the vessel are N. Initially
the piston is held against the pressure pi of the
gas with the help of two pins. In this state
temperature of the gas is T. Atmospheric pressure
is po.
,d yEck csyukdkj ik= iw.kZr;k pkyd nhokjksa ls
cuk gSA bl ik= esa ,d vkn'kZ ,dijekf.od xl
S
,d gYds fiLVu dh lgk;rk ls Hkjh gS rFkk ;g fiLVu
csyu ds vUnj fcuk ?k"kZ.k xfr dj ldrk gAS bl ik=
esa xl
S ds ijek.kqvksa dh la [;k N gAS izkjEHk esa bl
fiLVu dks nks fiuksa dh lgk;rk ls xl
S ds pi nkc ds
fo:¼ jksddj j[kk x;k gAS bl voLFkk esa xl
S dk
rkieku T gS o ok;qe.Myh; nkc po gAS
A
LL
EN
33.
2014
When the pins are removed and the holes are
closed, the gas expands rapidly and finally the
piston settles to a position where the force on
the piston due to the gas pressure balances the
force due to the atmospheric pressure. Which
of the following is a correct expression for the
work done by the gas during its rapid
expansion?
(1) Zero
fiuksa dks gVkus ij rFkk fNæ dks cUn dj nsus ij] xl
S
'kh?kzrk ls izlkfjr gksrh gS vkjS var esa fiLVu ml fLFkfr
esa vk tkrk gS tgk¡ xl
S ds nkc ds dkj.k fiLVu ij
yxus okyk cy] ok;qe.Myh ; nkc ds dkj.k yxus
okys cy ds cjkcj gks tkrk gAS xl
S ds bl rhoz izlkj
ds nkSjku xl
S }kjk fd;k x;k dk;Z gksxk %&
(1) 'kwU;
æ p0 ö
(2) NkT ln ç ÷
è p1 ø
æ p0 ö
(2) NkT ln ç p ÷
è 1ø
æ p ö
(3) NkT ç 1 - 0 ÷
p1 ø
è
(4) Information is insufficient to decide
æ p ö
(3) NkT ç 1 - 0 ÷
p1 ø
è
(4) vk¡dM+s vi;kZIr gAS
H-10/31
jQ dk;Z ds fy;s txg
ALLEN
35.
36.
37.
A
LL
EN
34.
2014
The K, L and M energy levels of platinum lie 34. IysfVue ds K, L o M ÅtkZ Lrj Øe'k% 78, 12 o
roughly at 78, 12 and 3 keV respectively. The
3 keV ij gksrs gaSA X-fdj.k LiDS Vªe esa Ka o Kb js[kk
ratio of wavelength of Ka line to that of Kb
dh rjaxn/S ;Z dk vuqikr gksxk %&
line in X-ray spectrum is22
3
22
25
22
3
22
25
(1)
(2)
(3)
(4)
(1)
(2)
(3)
(4)
3
22
25
22
3
22
25
22
A red star and a green star radiate energy at 35. ,d yky rkjk o ,d gjk rkjk leku nj ls ÅtkZ
the same rate which star is bigger in size.
fofdfjr djrs gaSA fdl rkjs dk vkdkj cM+k gS %&
(1) red
(1) yky
(2) green
(2) gjk
(3) both have same size
(3) nksuksa dk vkdkj leku gAS
(4) Can't say anything
(4) dqN ugha dg ldrsA
For a prism its refractive index is cot A/2 then 36. ,d fizTe ds fy, viorZukad cot A/2 g]S rks U;wure
minimum angle of deviation is :fopyu dks.k dk eku D;k gksxk %&
(1) 180 – A
(2) 180 – 2A
(1) 180 – A
(2) 180 – 2A
(3) 90 – A
(4) A/2
(3) 90 – A
(4) A/2
Two identical coaxial rings each of radius R 37. nks ,d tSlh lek{kh; oy;ksa esa ls izR;sd dh f=T;k
are separated by a distance of 3R . They are
uniformly charged with charges +Q and –Q
respectively. The minimum kinetic energy with
which a charged particle (charge +q) should be
projected from the center of the negatively
charged ring along the axis of the rings such
that it reaches the center of the positively
charged ring is :Qq
Qq
(1)
(2)
4 pe0 R
2pe 0 R
(3)
Qq
8pe0 R
(4)
3Qq
4 pe0 R
ALLEN
R gS rFkk ;s ,d&nwljs ls
3R nwjh ij j[kh gqbZ gSaA
bUgsa Øe'k% +Q rFkk –Q vkos'kksa }kjk ,d leku :i
ls vkosf'kr fd;k tkrk gSA ,d vkosf'kr d.k (vkos'k
+q) dks ½.kkosf'kr oy; ds dsUæ ls oy; dh v{k
ds vuqfn'k fdl U;wure xfrt ÅtkZ ds lkFk iz{ksfir
fd;k tk, rkfd ;g /kukosf'kr oy; ds dsUæ rd igq¡p
lds :-
(1)
Qq
4 pe0 R
(2)
Qq
2pe 0 R
(3)
Qq
8pe0 R
(4)
3Qq
4 pe0 R
jQ dk;Z ds fy;s txg
H-11/31
2014
38.
Two large parallel planes charged uniformly
with surface charge density s and –s are
located as shown in the figure. Which one of
the following graphs shows the variation of
electric field along a line perpendicular to the
planes as one moves from A to B ?
s
A
LL
EN
(4)
x
x
A charged particle enters into a uniform
magnetic field with velocity v0 perpendicular
39.
3
R,
2
where R is the radius of the circular path of the
particle in the field. The magnitude of change
in velocity of the particle when it comes out of
the field is :-
to it, the length of magnetic field is x =
(3)
x
E
(3)
x
3v 0
2
(4) v0
H-12/31
x
E
(4)
v0
2
(2)
x
E
(2)
E
(1)
(2)
(1) 2v0
B
E
E
E
–s
A
x
39.
s
B
E
(3)
fp= esa iznf'kZr nks yEcs lekUrj lery ] i`"Bvkos'k
?kuRo s rFkk –s }kjk ,d leku :i ls vkosf'kr gSaA
A ls B dh vksj tkus ij bu leryksa ds yEcor~ js[kk
ds vuqfn'k fo|qr {ks= esa ifjorZu dks n'kkZus okyk vkjs[k
gksxk :-
–s
A
(1)
38.
x
,d vkosf'kr d.k v0 osx ls ,d le:i pqEcdh;
{ks= esa blds yEcor~ izo's k djrk gAS pqEcdh; {ks=
dh yEckbZ x =
3
R g]S tgk¡ R bl {ks= esa d.k ds
2
o`Ùkkdkj iFk dh f=T;k gSA tc d.k {ks= ls ckgj
fudyrk gS rks blds osx es a ifjorZ u dk ifjek.k
gksxk %&
(1) 2v0
(2)
v0
2
(3)
3v 0
2
(4) v0
jQ dk;Z ds fy;s txg
ALLEN
2014
40.
A piece of conducting wire of resistance R is 40.
cut into 2n equal parts. Half the parts are
connected in series to form a bundle and
remaining half in parallel to form another
bundle. These bundles are then connected to
give the maximum resistance. The resistance of
the combination is
(3)
41.
R
1 + n2
2
Ræ
1 ö
ç1 + 2 ÷
2è
n ø
(2)
R
2(1 + n 2 )
1ö
æ
(4) R ç n + ÷
nø
è
(
)
In the circuit shown, when the switch S is
closed-
41.
(1) no charge flows through S
(2) charge flows from A to B
(3) charge flows from B to A
(4) charge flows initially from A to B and later
from B to A
ALLEN
R
1 + n2
2
(1)
Ræ
1 ö
ç1 + 2 ÷
2è
n ø
(2)
(3)
R
2(1 + n 2 )
1ö
æ
(4) R ç n + ÷
nø
è
A
LL
EN
(1)
izfrjks/k R okys ,d pkyd rkj ds VqdM+s dks leku
2n Hkkxksa esa dkVk tkrk gSA buesa ls vk/ks VqdM+ksa dks
Js.khØe esa tksM+dj ,d c.My cuk;k tkrk gS rFkk
'ks"k vk/ks Hkkxksa dks lekUrj Øe esa tksM+dj ,d vU;
c.My cuk;k tkrk gAS vc bu c.Myksa dks vkil
esa bl izdkj tksM+k tkrk gS rkfd vfèkdre izfrjks/k
izkIr fd;k tk ldsA bl la;kstu dk izf rjks/k gksxk
(
)
iznf'kZr ifjiFk esa ] tc fLop S cUn fd;k tkrk g&S
(1) S ls dksbZ vkos'k izokfgr ugha gksrk gAS
(2) A ls B dh vksj vkos'k izokfgr gksrk gAS
(3) B ls A dh vksj vkos'k izokfgr gksrk gAS
(4) izkjEHk esa vkos'k A ls B dh vksj o ckn esa B ls
A dh vksj izokfgr gksrk gSA
jQ dk;Z ds fy;s txg
H-13/31
2014
42.
A uniform conducting rectangular loop of sides 42.
l, b and mass m carrying current i is hanging
horizontally with the help of two vertical strings.
There exists a uniform horizontal magnetic field
B which is parallel to the longer side of loop. The
value of tension which is least is
,d le:i pkyd vk;rkdkj ywi dh Hkqtk,a l o b
gaS rFkk bldk nzO;eku m gAS blesa i /kkjk izokfgr gks
jgh gAS bls nks ÅèokZ/kj jfLl;ksa dh lgk;rk ls {kfS rt
:i ls yVdk j[kk gAS ;gka ,d le:i {kSfrt pqEcdh;
{ks= B fo|eku gS tks fd ywi dh cM+h Hkqtk ds lekUrj
gSA jfLl;ksa esa ruko dk U;wure eku gksxk
b
A
LL
EN
b
B
l
(1) mg – Bbi
(3)
43.
(2) mg + Bbi
mg - 2Bbi
2
(4)
(1) mg – Bbi
mg + 2Bbi
2
(3)
An infinitely large nonconducting plane of 43.
uniform surface charge density s has circular
aperture of certain radius R carved out from
it. The electric field at a point which is at a
distance ‘a’ from the centre of the aperture and
perpendicular to the plane is
s
2 2e 0
B
l
. The
(2) mg + Bbi
mg - 2Bbi
2
(4)
,dleku i`"B vkos'k ?kuRo s okys ,d vuUr yEcs
vpkyd ry esa ls ,d fuf'pr f=T;k R okys o`Ùkh;
}kjd dks dkVk x;k gSA ry ds yEcor~ rFkk }kjd
ds dsUnz ls a nwjh ij fLFkr ,d fcUnq ij fo|qr {ks=
dk eku
s
2 2e 0
gAS }kjd dh f=T;k R dk eku gksxk
radius of aperture R is
(1) a
(2)
2a
a
(3)
2
(4) 2a
H-14/31
mg + 2Bbi
2
(1) a
(2)
2a
(3)
a
2
(4) 2a
jQ dk;Z ds fy;s txg
ALLEN
2014
44.
A non-planar circular loop consists of two 44.
semi-circles each of radius R, one of which lies
in yz-plane & the other is in xz-plane as shown
and carries a current I. The magnetic force
experienced by positive charge of value Q
moving with velocity v along x direction when
it is at the origin is :
y
,d vleryh; o`Ùkkdkj ywi fp=kuqlkj izR;sd R
f=T;kvksa okys nks v¼Zo`Ùkksa ls feydj cuk gS ] ftuesa
ls ,d yz-ry esa rFkk nwljk xz-ry esa fLFkr gS rFkk
blesa I èkkjk izokfgr gks jgh gAS x fn'kk ds vuqfn'k v
osx ls xfr'khy Q /kukos'k tc ewy fcUnq ij gS rks bl
ij yxus okyk pqEcdh; cy gksxk%&
y
I
I
x
A
LL
EN
x
z
z
(1)
Qvm0 I
4R
(2)
Qvm0 I
2R
Qvm0 I
(4) 0
2 2R
A rectangular loop PQRS, is being pulled with 45.
constant speed into a uniform transverse
magnetic field by a force F (as shown). E.m.f.
induced in side PS and potential difference
between points P and S respectively are
(Resistance of the loop = r)
(3)
45.
Fr
Bl
(2) zero, Zero
(4)
Fr
Fr
,
6Bl 6Bl
×
P
l
S
2l
×
×Q
×
×
×
×
×R
×
×
× F ×
×
×
×
B
×
(2)
(3)
Qvm0 I
2 2R
(4) 0
Qvm0 I
2R
fdlh vk;rkdkj ywi PQRS dks fp=kuqlkj cy F
}kjk le:i vuqizLFk pqEcdh; {ks= esa fu;r pky ls
[khapk tkrk gSA Hkqtk PS esa izsfjr fo|qr okgd cy
rFkk fcUnqvksa P o S ds e /; foHkokUrj Øe'k % gS
(ywi dk izfrjksèk = r)
Fr
Bl
(2) 'kwU;, 'kwU;
Fr
(3) 'kwU;,
6Bl
(4)
ALLEN
Qvm0 I
4R
(1) 'kwU;,
(1) Zero,
Fr
(3) Zero,
6Bl
(1)
×
P
l
S
Fr
Fr
,
6Bl 6Bl
2l
×
×Q
×
×
×
×
× F ×
×
×
×
×
×R
×
B
jQ dk;Z ds fy;s txg
H-15/31
×
2014
47.
48.
An open capillary tube is lowered in a vessel 46.
with mercury. The difference between the
levels of the mercury in the vessel and in the
capillary tube Dh = 4.6mm. What is the radius
of curvature of the mercury meniscus in the
capillary tube? Surface tension of mercury is
0.46 N/m, density of mercury is 13.6 gm/cc.
1
1
m
(2)
m
(1)
340
680
1
(3)
m
(4) Information insufficient
1020
In a series L.C.R. a.c. circuit at off–resonance, 47.
the value of the angular frequency for which
the some voltage leads the current in the circuit,
is :1
1
(1) w <
(2) w >
LC
LC
1
(3) w =
(4) None of these
LC
A uniform rod hinged at its one end is allowed 48.
to rotate in vertical plane. Rod is given an
angular velocity w in its vertical position as
shown in figure. The value of w for which the
force exerted by the hinge on rod is zero in this
position is–
(1)
(3)
,d [kqyh ds'kuyh dks ikjs ls Hkjs ik= esa Mqcks;k tkrk
gAS ik= rFkk ds'kuyh esa ikjs ds Lrjksa ds e/; vUrj
Dh = 4.6mm gSA ds' kuyh esa cuus okys ikjs ds
uopUæd dh oØrk f=T;k D;k gksxh tcfd ikjs dk
i`"Bruko 0.46 N/m o ikjs dk ?kuRo 13.6 gm/cc
gSA
1
1
m
(2)
m
340
680
1
m
(4) vk¡dM+s vi;kZIr gAS
(3)
1020
Ükà[kykc¼ L.C.R. izR;korhZ /kkjk ifjiFk dh vuqukn
(1)
A
LL
EN
46.
g
L
(2)
g
2L
(4)
2g
L
(1) w <
1
LC
(2) w >
(3) w =
1
LC
(4) buesa ls dksbZ ugha
(1)
L
3g
L
1
LC
,d leku NM+] ,d fljs ij dhydhr gS rFkk Å/okZèkj
ry esa ?kweus ds fy, Lora= gAS NM+ dks bldh ÅèokZèkj
fLFkfr ls fp=kuqlkj dks.kh; osx w nsrs gaSA w dk eku
ftl ds fy;s bl fLFkfr esa dhyd }kjk NM+ ij yxk;k
cy 'kwU; g]S gksxk&
w
H-16/31
vUR; fLFkfr esa dks.kh; vko`fÙk ds fdl eku ds fy;s
lzksr dh oksYVrk ifjiFk dh /kkjk ls vxz gksxh\
(3)
g
L
(2)
g
2L
(4)
2g
L
w
L
3g
L
jQ dk;Z ds fy;s txg
ALLEN
50.
A
LL
EN
49.
2014
A star is modeled as a uniform spherical 49. ,d rkjs dk foU;kl bl izdkj gS fd tSls inkFkZ dk
distribution of matter. How gravitational
le:i xksyh; forj.k gksA bldh lrg ij xq:Rokd"kZ.k
pressure on surface depends on volume of the
nkc] rkjs ds vk;ru ij fdl izdkj fuHkZj djrk g\
S
star?
–1/3
–1/3
(1) P µ V
(2) P µ V
(1) P µ V
(2) P µ V
–2/3
–2/3
–4/3
(3)
P
µ
V
(4)
P µ V–4/3
(3) P µ V
(4) P µ V
A transverse wave is propagating along +x 50. ,d vuqizLFk rjax +x fn'kk ds vuqfn'k lapfjr gks jgh
g SA t = 2 sec ij x = 4m ij fLFkr d.k
direction. At t = 2 sec, the particle at x = 4m
y
= 2 mm ij gAS le; xqtjus ds lkFk bldk y funsZ'kkad
is at y = 2 mm. With the passage of time its y
c<+rk gS rFkk vfèkdre 4 mm rd igq¡p tkrk gAS rjax
coordinate increases and reaches to a
lehdj.k gks ldrh gS (;gk¡ w o k ds lkekU; vFkZ
maximum of 4 mm. The wave equation may
g
AS )%&
be (using w and k with their usual meanings)
p
p
(1) y = 4 sin(w(t + 2) + k(x - 2) + )
(1) y = 4 sin(w(t + 2) + k(x - 2) + )
6
6
p
(2) y = 4 sin(w(t + 2) + k(x) + )
6
p
(2) y = 4 sin(w(t + 2) + k(x) + )
6
(3) y = 4 sin(w(t - 2) - k(x - 4) +
51.
5p
)
6
(3) y = 4 sin(w(t - 2) - k(x - 4) +
5p
)
6
p
(4) y = 4 sin(w(t - 2) - k(x - 4) + )
6
Two tuning forks A and B produce 8 beats/s 51.
when sounded together. A gas column 37.5 cm
long in a pipe closed at one end resonate to its
fundamental mode with fork A whereas a
column of length 38.5 cm of the same gas in
a similar pipe is required for a similar resonance
with fork B. The frequencies of these two
tuning forks, are :-
p
(4) y = 4 sin(w(t - 2) - k(x - 4) + )
6
A vkjS B nks Lofj= f}Hkqt lkFk&lkFk Lofjr gksus ij
çfr lsd.M 8 foLian mRiUu djrs gSaA ,d fljs ls
cUn ikbZi esa ,d xl
S dk 37.5 lseh yEck LrEHk viuh
ewy fo/kk esa Lofj = A ls vuqukfnr gksrk gS tcfd
Lofj= B blh xl
S ds ,sls gh ikbZi esa 38.5 lseh LrEHk
(1) 308 Hz, 300 Hz
(2) 208 Hz, 200 Hz
(3) 300 Hz, 400 Hz
(4) 350 Hz, 500 Hz
(1) 308 Hz, 300 Hz
(3) 300 Hz, 400 Hz
jQ dk;Z ds fy;s txg
ALLEN
ls vuqukfnr gksrk gAS nksuksa Lofj= f}Hkqtksa dh vko`fÙk;k¡
gksxh %&
(2) 208 Hz, 200 Hz
(4) 350 Hz, 500 Hz
H-17/31
2014
53.
Parallel rays are incident on a thick plano-convex 52.
lens having radius of curvature R, refractive
index µ and thickness t. When rays are incident
on plane surface they converge at a distance x
from plane surface. When rays are incident on
curved surface then rays converge at y distance
from curved surface. Then
,d eks Vs leryksÙ ky ysU l dh oØrk f=T;k R,
viorZukad µ o eksVkbZ t gAS bl ysUl ij lekUrj fdj.ksa
vkifrr gksrh gAS tc ;s lery lrg ij vkifrr gksrh
gS rks lery lrg ls x nwjh ij vfHklfjr gks tkrh
gSA tc ;s oØh; lrg ij fxjrh gS rks oØh; lrg
ls y nwjh ij vfHklfjr gksrh gSA rc%&
(1) x = y
(2) x < y
(1) x = y
(2) x < y
(3) x > y
(4) data insufficient
(3) x > y
(4) vkadM+s vi;kZIr gAS
A
LL
EN
52.
Three alternating voltage sources V1 = 3 sinwt
volt, V2= 5 sin(wt + f1) volt and V3 = 5 sin(wt – f2)
volt connected across a resistance R =
53.
7
W as
3
shown in the figure (where f 1 and f 2
corresponds to 30° and 127° respectively). Find
the peak current (in Amp) through the resistor.
rhu iz R;korhZ oksY Vrk lz k s r V 1 = 3sinwt oks Y V ,
V2=5sin(wt + f1) oksYV rFkk V3 = 5 sin(wt – f2)
volt dks fp=kuqlkj R =
tksM+k x;k gAS (;gka f1 rFkk f2 ds eku Øe'k% 30° o
127° g)
S izfrjks/kd ls izokfgr f'k[kj /kkjk (Amp es)a
Kkr dhft,A
V3
V3
V1
(1) 3
(2) 4
(3) 5
(4) 6
H-18/31
Ö7/3W
V2
Ö7/3W
V2
7
W izfrjks/k ds fljksa ij
3
(1) 3
V1
(2) 4
(3) 5
(4) 6
jQ dk;Z ds fy;s txg
ALLEN
2014
54.
55.
A
54.
foLFkkiu /kkjk %&
(1) eqä bysDVªkWuksa ds izokg ds dkj.k mRiUu pkyu
/kkjk ds leku gksrh gS
(2) /kuk;uksa ds izokg ds dkj.k mRiUu pkyu /kkjk ds
leku gksrh gSA
(3) /kukRed rFkk ½.kkRed nksuksa izdkj ds eqä vkos'k
okgdksa ds izokg ds dkj.k mRiUu pkyu /kkjk ds
leku gksrh gSA
(4) ;g pkyu /kkjk ugha gksrh cfYd ;g le; ifjorhZ
fo|qr {ks= ds dkj.k mRiUu gksrh gAS
55. fn, x, ifjiFk fp= ds fy, lR;rk lkj.kh gksxh%&
A
LL
EN
Displacement current
(1) is same as conduction current due to flow
of free electron
of positive ions
of both positive and negative free charge
carriers
(4) is not a conduction current but is caused by
time varying electric field
Truth table for the given circuit figure is :C
A
C
E
B
D
A
B
E
(1)
0
0
1
1
0
1
0
1
1
0
1
0
(3)
0
0
1
1
0
1
0
1
0
1
0
1
B
A
B
E
(2)
0
0
1
1
0
1
0
1
1
0
0
1
(4)
0
0
1
1
0
1
0
1
0
1
1
0
ALLEN
E
D
A
B
E
(1)
0
0
1
1
0
1
0
1
1
0
1
0
(3)
0
0
1
1
0
1
0
1
0
1
0
1
A
B
E
(2)
0
0
1
1
0
1
0
1
1
0
0
1
(4)
0
0
1
1
0
1
0
1
0
1
1
0
jQ dk;Z ds fy;s txg
H-19/31
2014
56.
An electron (mass m) with an initial velocity
56.
v = n 0 î is in an electric field E = E0 ˆj . If
v = n0 î gS rFkk ;g fo|qr {ks= E = E0 ˆj esa gAS ;fn
l0 = h/mn0, it's de-Brogile wavelength at time
t is given by
(2) l 0
(1) l0
l0
1+
57.
2
(4)
2 2
0
2 2
0
e Et
m n
gksxh %&
e 2 E 20 t 2
1+ 2 2
m n0
(2) l 0 1 +
(1) l0
l0
æ e 2 E 20 t 2 ö
ç1 + 2 2 ÷
m n0 ø
è
Poynting vector (which gives the direction of
electromagnetic waves) is defined as :
r r r
r r r
(2) J = E.B
(1) J = E × B
r r
r E×B
(3) J =
2
58.
l0 = h/mn0 gks rks le; t ij bldh Mh&czkXs yh rjaxnèS ;Z
l0
A
LL
EN
(3)
,d m æO;eku okys bys D Vª k W u dk iz k jfEHkd os x
(3)
57.
r r r r r
(4) J = E × B + E.B
For a paramagnetic material, the dependence of 58.
the magnetic susceptibility cm on the absolute
temperature T is given by :
(1) cm µ T
(2) cm µ exp (constant × T)
2
2 2
0
2 2
0
e Et
1+
m n
æ e 2 E 20 t 2 ö
ç1 + 2 2 ÷
m n0 ø
è
r r r
(1) J = E × B
r r r
(2) J = E.B
r r
r E×B
(3) J =
2
r r r r r
(4) J = E × B + E.B
,d vuqpqEcdh; inkFkZ ds fy, pqEcdh; lqxzkfgrk cm
dh ije rki T ij fuHkZjrk nh tkrh gS %&
(1) cm µ T
(2) cm µ exp (constant × T)
(3) cm µ (1/T)
(4) cm = constant
(4) cm = fu;r
H-20/31
l0
fo|qr pqEcdh; rjaxksa dh fn'kk n'kkZus okyk Poynting
lfn'k fy[kk tkrk gS %&
(3) cm µ (1/T)
(4)
e 2 E 20 t 2
m 2 n 02
jQ dk;Z ds fy;s txg
ALLEN
60.
Statement 1 : When a closed organ pipe 59.
vibrates, the pressure of the gas at the closed
end remains constant.
and
Statement–2 : In a stationary–wave system,
displacement nodes are pressure antinodes, and
displacement antinodes are pressure nodes.
(1) Statement–1 is True, Statement–2 is True;
Statement–2 is a correct explanation for
Statement–1.
Statement–2 is not a correct explanation for
Statement–1.
(3) Statement–1 is True, Statement–2 is False.
(4) Statement–1 is False, Statement–2 is True.
Statement 1 : When a bottle of cold carbonated 60.
drink is opened, a slight fog forms around the
opening.
and
Statement 2 : Adiabatic expansion of the gas
causes lowering of temperature which starts
condensation of water vapours.
Statement–2 is a correct explanation for
Statement–1.
Statement–2 is not a correct explanation for
Statement–1.
(3) Statement–1 is True, Statement–2 is False.
(4) Statement–1 is False, Statement–2 is True.
oDrO;–1: tc ,d cUn vkW xZu ikbi dEiUu djrk
gS rks cUn fljs ij xSl dk nkc fu;r jgrk gSA
vkS j
oDrO;–2 : fdlh vizxkeh rjax fudk; esa foLFkkiu
fuLiUn] nkc izLiUn gksrs gaS rFkk foLFkkiu izLiUn] nkc
fuLiUn gksrs gSaA
(1) oäO;&1 lR; g]S oäO;&2 lR; g]S oäO;&2]
oäO;&1 dk lgh Li"Vhdj.k gAS
(2) oäO;&1 lR; g]S oäO;&2 lR; gS ; oäO;&2]
oäO;&1 dk lgh Li"Vhdj.k ugha gAS
(3) oäO;&1 lR; g,S oäO;&2 vlR; gAS
A
LL
EN
59.
2014
ALLEN
(4) oäO;&1 vlR; g]S oäO;&2 lR; gAS
oDrO;–1: tc B.Ms dkcksZuVs M
s is; ls Hkjh ,d cksry
dks [kksyk tkrk gS rks cksry ds eq¡g ij pkjksa vksj gYds
ls >kx cu tkrs gaSA
vkS j
oDrO;–2 : xl
S ds :¼ks"e izlkj ds dkj.k rkieku
?kV tkrk gS ftlls ty ok"i dk la?kuu izkjEHk gks tkrk
gSA
(1) oäO;&1 lR; g]S oäO;&2 lR; g]S oäO;&2]
oäO;&1 dk lgh Li"Vhdj.k gAS
(2) oäO;&1 lR; g]S oäO;&2 lR; gS ; oäO;&2]
oäO;&1 dk lgh Li"Vhdj.k ugha gAS
(3) oäO;&1 lR; g,S oäO;&2 vlR; gAS
(4) oäO;&1 vlR; g]S oäO;&2 lR; gAS
jQ dk;Z ds fy;s txg
H-21/31
2014
PART C - CHEMISTRY
62.
63.
e.m.f. of the cell
2Ag+ + Cu ® Cu+2 + 2Ag
Given : EºAg/Ag+ = –0.8 V ; EºCu+2/Cu = 0.3 V
(3)
lsy dk e.m.f. gksxk&
2Ag+ + Cu ® Cu+2 + 2Ag
fn;k gS % EºAg/Ag+ = –0.8 V ; EºCu+2/Cu = 0.3 V
(1) –0.5 V (2) 0.5 V (3) –1.1 V (4) 1.1 V
pH of a solution obtained by mixing equal 62.
volume of 0.2 M NaOH & 0.2 M CH3COOH
(Ka = 10–5) is :
(1) 7
(2) 5
(3) 9
(4) 9.5
+
If shortest wavelength of He ion in Balmer 63.
series is X metres then longest wavelength in
Paschen series of Li+2 ion is :
(1)
64.
61.
(1) –0.5 V (2) 0.5 V (3) –1.1 V (4) 1.1 V
0.2 M NaOH rFkk 0.2 M CH3COOH ds leku
vk;ru dks feykus ij izk Ir foy;u dh pH D;k
gksxh\(Ka = 10–5)
(1) 7
(2) 5
(3) 9
(4) 9.5
+
;fn ckej Js.kh esa He vk;u dh y?kq re rjaxnS/;Z
X ehVj gS rks Li+2 vk;u dh ik'pu Js.kh esa nh?kZre
A
LL
EN
61.
rjaxn/S ;Z gksxh &
36
X
5
(2)
16
X
7
(1)
36
X
5
(2)
16
X
7
9
X
5
(4)
5
X
9
(3)
9
X
5
(4)
5
X
9
Which one of the following graph is incorrect
depiction of first order process "A ® P"
64.
fuEu oØksa esa ls dkSulk oØ izFke dksfV izØe
"A ® P" dk xyr izn'kZu gAS
(1)
(1)
(2)
log[A]
(2)
log[A]
log [P]
t
t
t
t
[P]
[A]
(3)
rate of
disappearance
of [A]
(4)
t
(3) ds foyksiu
rate of
appearance
of [P]
dh nj
t
H-22/31
log [P]
(4) ds cuus
t
dh nj
t
jQ dk;Z ds fy;s txg
ALLEN
2014
65.
Calculate the number of milli moles of SO2. If
in the following reaction 10 ml of 0.1 M
KMnO4 solution are required for titration.
65.
–2
–2
67.
SO2 + MnO¯
® SO4 + Mn+2
4
SO2 + MnO¯
® SO4 + Mn+2
4
(1) 2.5
(2) 0.5
(1) 2.5
(2) 0.5
(3) 1.25
(4) None of these
(3) 1.25
Which of the following reactions is
spontaneous only at relatively low temperature-
66.
fuEu vfHkfØ;kvksa esa ls dkuS lh dsoy lkis{k U;wu rki
ij Lor% gksrh gS -
A
LL
EN
66.
;fn fuEu vfHkfØ;k esa vuqekiu ds fy , 0.1 M
KMnO4 foy;u ds 10 ml dh vko';drk gksrh gS
rks SO2 ds fefy eksyksa dh la[;k Kkr dhft,A
(1) NH4Br(s) + 188 kJ ® NH3(g) + Br2(l)
(1) NH4Br(s) + 188 kJ ® NH3(g) + Br2(l)
(2) NH3(g) + HCl(g) ® NH4Cl(s) + 176kJ
(2) NH3(g) + HCl(g) ® NH4Cl(s) + 176kJ
(3) 2H2O2(l) ® 2H2O(l) + O2(g) + 196 kJ
(3) 2H2O2(l) ® 2H2O(l) + O2(g) + 196 kJ
(4) Both (2) & (3)
(4) (2) o (3) nksuksa
Select the incorrect statement:-
67.
xyr dFku dk p;u dhft,s&
(1) Stoichiometry of crystal remains uneffected
due to schottky defect
(1) fØLVy dh jllehdj.kferh] 'kkWV dh =qfV ds
(2) Frenkel defect usually shown by ionic
compound having low coordination
number
(2) Ýsady =qfV] mu vk;fud ;kSfxdksa }kjk iznf'kZr
(3) F-centres generation is responsible factor
for imparting the colour to the crystal
(3) F-dsUæksa dk fuekZ.k fØLVy ds jax ds fy, mÙkjnk;h
(4) Density of crystal always increases due to
substitutional impurity defect
(4) fØLVy dk ?kuRo loZnk izfrLFkkiuh; v'kqf¼ =qfV
ALLEN
dkj.k vizHkkfor jgrh gAS
dh tkrh gS] ftuesa lkekU;r% leUo;u la[;k de
gksrh gS
gksrs gaAS
ds dkj.k c<rk gSA
jQ dk;Z ds fy;s txg
H-23/31
2014
69.
Arrange the following aqueous solutions in
order of increasing freezing points (that is
lowest first) :-
I. 0.10 m Ba3(PO4)2 II. 0.10 m Na2SO4
III. 0.10 m C2H5OH
IV. 0.10 m KCl
V. 0.10 m Li3PO4
(1) I < V < II < IV < III
IV. 0.10 m KCl
V. 0.10 m Li3PO4
(1) I < V < II < IV < III
(2) III º IV < II < V < I
(2) III º IV < II < V < I
(3) IV < II < V < I < III
(3) IV < II < V < I < III
(4) I < V < II < IV = III
(4) I < V < II < IV = III
How many litres of oxygen at 1 atm &
273 K will be required to burn completely
2.2 g of propane (C3H8) :-
69.
(1) 11.2 L
(3) 5.6 L
(2) 22.4 L
(4) 44.8 L
(2) 3 atm
(3) 0.3 atm
(4) 0.18 atm
Which of the following complex is
paramagnetic in nature.
(1) K2[NiF6]
71.
(2) [Co(H2O)6]+3
(2) 22.4 L
(4) 44.8 L
1000 K rFkk 0.5 atm ds nkc ij ,d ik= esa CO2
xl
S Hkjh xbZ gAS xzQ
s kbV dks feykus ij CO2 dk dqN
Hkkx CO esa :ikUrfjr gks tkrk gS ;fn lkE; ij dqy
nkc 0.8 atm gks rks lkE; fu;rkad dk eku D;k gksxk\
(1) 1.8 atm
(2) 3 atm
(3) 0.3 atm
(4) 0.18 atm
fuEu esa ls dkSuls ladqy dh izd`f r vuqpqE cdh ;
gS -
(1) K2[NiF6]
(2) [Co(H2O)6]+3
II
II
(3) K4[ Fe (CN)5(O2)] (4) None of these
H-24/31
1 atm rFkk 273 K ij 2.2 g izkis us (C3H8) dks iw.kZr%
tykus ds fy, fdrus yhVj vkWDlhtu dh vko';drk
gksrh gaS&
A vessel at 1000 K contains CO 2 with a 70.
pressure of 0.5 atm. Some of the CO2 is
converted into CO on the addition of graphite.
The value of equilibrium constant if the total
pressure at equilibrium is 0.8 atm is:
(1) 1.8 atm
71.
fuEu tyh; foy;uksa dks fgekad fcUnq esa o`f¼ ds Øe
esa O;ofLFkr dhft;s & (vFkkZr~ U;wuere dks igys j[krs
gq,)s
I. 0.10 m Ba3(PO4)2 II. 0.10 m Na2SO4
III. 0.10 m C2H5OH
(1) 11.2 L
(3) 5.6 L
70.
68.
A
LL
EN
68.
(3) K4[ Fe (CN)5(O2)] (4) buesa ls dksbZ ugha
jQ dk;Z ds fy;s txg
ALLEN
72.
73.
Which of the following compound has 72.
3c – 4e– bond
(1) AlCl3
(2) BeCl2
(3) Both (1) and (2) (4) None of these
Select which of the following reaction is not 73.
correct (only product wise)(1) Mg3N2 + H2O ® Mg(OH)2 + NH3
(2) Mg2C3+H2O®Mg(OH)2 +CH3–CºCH
(4) Cu(NO3)2
Which of the following mixture is used for
making Holme's signal.
(1) Ca3P2 + Ca(OH)2 (2) Ca3P2 + CaC2
(3) CaC2 + PH3
76.
(1) AlCl3
(2) BeCl2
(3) (1) rFkk (2) nksuksa
ml vfHkfØ;k dk p;u dhft, tks lgh ugha gS (dsoy
mRikn ds fy,)
D
® Al(OH)3+CH3–CºCH
(3) Al4C3+H2O¾¾
A
LL
EN
D
® MgO + CO2
(4) MgCO3 ¾¾
Which of the following metal nitrate gives 74.
metal and oxygen on heating
(1) KNO 3
(2) AgNO3
(3) Be(NO3)2
75.
fdl ;kfS xd esa 3c – 4e– ca/k mifLFkr gS
(1) Mg3N2 + H2O ® Mg(OH)2 + NH3
(2) Mg2C3+H2O®Mg(OH)2 +CH3–CºCH
D
® Al(OH)3+CH3–CºCH
(3) Al4C3+H2O¾¾
74.
75.
(4) None of these
Which of the following statement is incorrect 76.
(1) When P4 reacts with SOCl2 then S2Cl2 is
formed
(2) When P4 reacts with SO2Cl2 then S2Cl2 is
formed
(3) Tin stone is diamagnetic in nature
(4) Bayer's process is used for the
concentration of red bauxite
ALLEN
2014
D
® MgO + CO2
(4) MgCO3 ¾¾
fuEu esa ls dkuS lh /kkrq ukbVªVs xeZ djus ij /kkrq rFkk
vkWDlhtu nsrh gS
(1) KNO 3
(3) Be(NO3)2
(2) AgNO3
(4) Cu(NO3)2
fuEu esa ls dkSulk feJ.k gkWYe~l ladrs (Holme's
signal) fuekZ.k ds fy, mi;ksx fd;k tkrk gS
(1) Ca3P2 + Ca(OH)2 (2) Ca3P2 + CaC2
(3) CaC2 + PH3
fuEu esa ls dkuS lk dFku xyr gS
(1) tc P4 ] SOCl2 ds lkFk vfHkfØ;k djrk gS rc
S2Cl2 dk fuekZ.k gksrk gS
(2) tc P4 ] SO2Cl2 ds lkFk vfHkfØ;k djrk gS rc
S2Cl2 dk fuekZ.k gksrk gS
(3) Vhu LVksu izfrpqEcdh; izd`fr dk gksrk gS
(4) cs;j izØe dk mi;ksx yky ckWDlkbM ds lkUnz.k
ds fy, fd;k tkrk gS
jQ dk;Z ds fy;s txg
H-25/31
2014
77.
Which of the following cation is colourless in 77.
fuEu esa ls dkSulk /kuk;u tyh; foy;u esa jaxghu
aq. solution
gksrk gSA
(1) Ni+2
79.
80.
81.
(1) Ni+2
(3) Co+3 (4) Ti+4
Select the correct statement :
78.
(2) Cr+3
(3) Co+3 (4) Ti+4
lgh dFku dk p;u dhft,
(1) O3 is diamagnetic in nature
(1) O3 izfrpqEcdh; izd`fr dh gksrh gS
(2) O3 has parmanent bleaching action
(2) O3 LFkk;h fojt
a u dk dk;Z djrk gS
(3) O3 is bent in shape
(3) O3 eqM+h gqbZ vkd`fr esa gksrh gS
(4) All of these
(4) mijksDr lHkh
Which of the following compound produces 79.
fuEu esa ls dk Sulk ;kSfxd ty vi?kVu djkus ij
H3PO4 on hydrolysis.
H3PO4 dk fuekZ.k djrk gS
A
LL
EN
78.
(2) Cr+3
(1) PCl5
(2) P4O10
(1) PCl5
(2) P4O10
(3) POCl3
(4) All of these
(3) POCl3
(4) mijksDr lHkh
Which of the following species is diamagnetic 80.
dkSulh Lih'kht izfrpqEcdh; gksrh gS
(1) O2+2
(2) N2
(1) O2+2
(2) N2
(3) Na+
(4) All of these
(3) Na+
(4) mijksDr lHkh
Which compound does not react with NaHCO3 81.
dkuS lk ;kSfxd NaHCO3 ds lkFk vfHkfØ;k ugha djrk
but having chiral centre ?
gS ysfdu blesa fdjSy dsUnz mifLFkr gksrk gS ?
(1) Ascorbic acid
(2) Tartaric acid
(1) ,LdkWfcZd vEy
(2) VkVZfjd vEy
(3) Carbolic acid
(4) Citric acid
(3) dkcksZfyd vEy
(4) flfVªd vEy
H-26/31
jQ dk;Z ds fy;s txg
ALLEN
2014
82.
A(C10H12) [no chiral centre]
82.
A(C10H12) [dksbZ fdjsy dsUnz ugha ]
A gives white precipitate with ammoniacal
solution of AgNO3
A ] AgNO3 ds veksuhd`r foy;u ds lkFk 'osr
O3
NaOH
® B (C8H12O4) ¾¾¾
® C (C6H12)
A ¾¾¾
H2 O
+ CaO
NaOH
3
® B(C 8 H 12 O 4 ) ¾¾¾
® C(C 6 H 1 2 )
A ¾¾¾
H2 O
+ CaO
Cl2 / hn
¾¾¾
® only one monochloro derivative.
Cl2 / hn
¾¾¾
® dsoy ,d eksuksDyksjks O;q RiUuA A gS&
vo{ksi nsrk gS
O
A is
CºCH
(2) H2C=HC
CH=CH2
CºCH
(1) H3C–H2C
A
LL
EN
83.
(1) H3C–H2C
CH=CH2
(2) H2C=HC
(3) HCºC
CºCH
(3) HCºC
CºCH
(4) H2C=C
|
CH3
CH3
(4) H2C=C
|
CH3
CH3
1-Methyl cyclopentanol ® 2-methyl
cyclopentanol. To carry out following
conversion the sequence of reagent used is -
83.
(1) H2SO4 (Conc.) ; CH3Cl ; aq. KOH
(2) H2SO4 (Conc.) ; HgSO4 ; H2O, NaBH4
(3) H2SO4 (Conc.) ; B2H6 + H2O2, NaOH
(4) PCl3 ; CH3I ; H2O
ALLEN
1-esfFky lkbDyksiUs VsukWy® 2-esfFky lkbDyksiUs VsukWy
mijksDr :ikUrj.k dks izkIr djus ds fy, mi;ksx fd;s
x;s vfHkdeZdks dk Øe gS -
(1) H2SO4 (lkUæ) ; CH3Cl ; tyh; KOH
(2) H2SO4 (lkUæ) ; HgSO4 ; H2O, NaBH4
(3) H2SO4 (lkUæ) ; B2H6 + H2O2, NaOH
(4) PCl3 ; CH3I ; H2O
jQ dk;Z ds fy;s txg
H-27/31
2014
NO2
Br
84. CH2(CO2Et)2
NaNH2
(1)
NO2
NO2
B,B is 84. CH2(CO2Et)2
Cl
(2)
O 2N
Br
NaNH2
(1)
O2N
NO2
NO2
(4)
Br
Cl
(2)
O 2N
(3)
O 2N
NO2
(4)
Br
CH(CO2Et)2
85.
86.
The number of stereoisomers of 1, 3, 5-trichloro
1, 4-pentadiene is(1) 1
(2) 2
(3) 3
(4) 4
Major organic product (alkene) in the reaction is:
85.
86.
Br
Br
(2)
Me 3COK
¾¾¾¾
®
(leq.)
Br
Br
(1)
Br
(4)
H-28/31
CH(CO2Et)2
1, 3, 5- Vª k bZ D yks j ks 1, 4- is U VkMkbZ b u ds f=foe
leko;fo;ksa dh la [;k gS (1) 1
(2) 2
(3) 3
(4) 4
vfHkfØ;k esa eq [; dkcZfud mRikn (,Ydhu) gS :
Me 3COK
¾¾¾¾
®
(leq.)
Br
(3)
NO2
Br
Br
(1)
NO2
Cl
CH(CO2Et)2
NO2
B, B gS
CH(CO2Et)2
A
LL
EN
O 2N
NO2
Br
Cl
Cl
A
CH(CO2Et)2
NO2
CH(CO 2Et)2
CH(CO2Et)2
(3)
Br
Cl
A
CH(CO2Et)2
O2N
NO2
Br
(2)
Br
(3)
Br
Br
(4)
jQ dk;Z ds fy;s txg
ALLEN
2014
O
87.
O
+
H3O
¾¾¾¾¾¾
®
(complete hydrlysis)
O
organic 87.
product :
(1)
O
OH
(1)
O
O
OH
OH
OH
(2)
O
OH
OH
OH
HO
A
LL
EN
HO
O
(3)
O
(3)
OH
(4) None of these
88.
CHCl3 + KOH
X
(C7H9N)
NaNO2 + HCl
Bad smell
88.
CHCl3 + KOH
X
(C7H9N)
NaNO2 + HCl
Alcohol (not soluble in NaOH)
NH2
NH2
(1)
(2)
,YdksgkWy (NaOH esa foys; ugha)
X ] HCl esa foys; gAS X gksxk:
(2)
CH3
CH2–NH2
(4)
NH2
NH2
CH3
NH–CH3
nqxZU/k
(1)
CH3
ALLEN
OH
X is soluble in HCl. X will be :
(3)
dkcZ f ud
mRikn :
OH
(2)
+
H3 O
¾¾¾¾¾
®
(iw.kZ tyvi?kVu)
O
CH3
NH–CH3
(3)
CH2–NH2
(4)
jQ dk;Z ds fy;s txg
H-29/31
90.
An optically active monobasic acid having 89.
molecular weight 116 does not declourise
Baeyer's reagent. A on treatment with Br2/red
phosphorous produce B which is still optically
active. B on dehydro bromination followed by
decarboxylation by (NaOH + CaO) gives
major product C which does not show
stereoisomerism. Structure of A can be :
CH3
|
(1) CH3CH2–CH–CH2CO2H
CH3
|
(2) CH3CH2–CH2–CH–CO2H
(3) CH3–CH–CH2–CH2–CO2H
|
CH3
CH3
|
(4) H3C–C—CH–CO2H
| |
CH3 CH3
CH3
|
(1) CH3CH2–CH–CH2CO2H
CH3
|
(2) CH3CH2–CH2–CH–CO2H
(3) CH3–CH–CH2–CH2–CO2H
|
CH3
CH3
|
H
C—
CH–CO2H
(4) 3C–
| |
CH3 CH3
Which of the following option is 90.
INCORRECT ?
(1) Amino acid is solid at isoelectric point
(2) Lactose is example of disaccharide &
reducing sugar
(3) Nylon 66 is condensation polymer & have
amide linkage
(4) In RNA molecules, the sugar moiety is
b-D-2 deoxyribose.
fuEu fodYiksa esa ls dkSulk xyr gS ?
(1) vehuks vEy leo|
S qr fcUnq ij Bksl gksrk gS
(2) ysDVksl] MkbZld
S sjkbM rFkk vipk;h 'kdZjk dk
mnkgj.k gksrk gS
(3) uk;ykWu 66 l?a kuu cgqyd gS vkjS blesa ,ekbM
ca/ku gksrk gS
(4) RNA v.kq esa 'kdZjk dk vk/kk Hkkx b-D-2
MhvkWDlhjkbckst gksrk gS
116 vkf.od Hkkj okyk ,d izdkf'kd lfØ; ,dy
{kkjh; vEy] cs;j vfHkdeZd dks jaxghu ugha dj
ldrkA A dh fØ;k Br2/yky QkLQksjl ds lkFk djkus
ij B izkIr gksrk gS tks izdkf'kd lfØ; gAS B dk
fogkbMª k s c z k s e huhdj.k djkus ds i'pkr~
(NaOH + CaO) }kjk fodkcksZDlhyhdj.k djkus ij
eq[; mRikn C nsrk g]S tks f=foe leko;ork ugha n'kkZrk
gSA A dh lajpuk gks ldrh gS :
A
LL
EN
89.
2014
H-30/31
jQ dk;Z ds fy;s txg
ALLEN
jQ dk;Z ds fy;s txg
A
LL
EN
2014
ALLEN
jQ dk;Z ds fy;s txg
H-31/31

ALLEN JEE-MAIN SAMPLE PAPER # 01 TARGET - 2014

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