[Type text] Page 1 JPT-1 (JEE ADVANCE) TEST DATE : | BATCH : JP,JF,JR FACULTY NAME: TARGET DATE : 16-06-2014 Syllabus : XI and XII S.No. Subject Nature of Questions No. of Questions Marks Negative Total 1 to 15 MCQ 15 4 0 60 16 to 19 Match matrix listing 4 3 1 12 20 to 34 MCQ 15 4 0 60 35 to 38 Match matrix listing 4 3 1 12 39 to 53 MCQ 15 4 0 60 54 to 57 Match matrix listing 4 3 1 12 57 216 Paper-2 Total Total 1. Let A = 0 0 0 and (A + ) 50 50A = a b c d then [MT-AL] [301] ekuk A = 0 0 0 vkSj (A + ) 50 50A = a b c d rc (A*) a + d = 2 (B*) a + b = 1 (C*) b + c = 0 (D) a + d = 1 Sol. A = 0 0 0 A 2 = 0 0 0 0 = 0 A 3 = 0 (A + ) 3 = A 3 + 3A 2 + 3A + Similarly blh izdkj (A + ) 50 = 50A + ......(1) (A + ) 50 50A = 1 0 0 1 ans. A,B,C a =1, b = 0, c = 0, d = 1 2. From a pack of 52 playing cards, all face cards are removed. From the remaining pack, cards are dealt one by one until an ace appear. Let P 1 denote the probability that exactly 10 cards are dealt before the first ace
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
[Type text] Page 1
JPT-1 (JEE ADVANCE) TEST DATE : | BATCH : JP,JF,JR
FACULTY NAME:
TARGET DATE : 16-06-2014
Syllabus : XI and XII
S.No. Subject Nature of Questions No. of Questions Marks Negative Total
1 to 15 MCQ 15 4 0 60
16 to 19 Match matrix listing 4 3 �1 12
20 to 34 MCQ 15 4 0 60
35 to 38 Match matrix listing 4 3 �1 12
39 to 53 MCQ 15 4 0 60
54 to 57 Match matrix listing 4 3 �1 12
57 216
Paper-2
TotalTotal
1. Let A =0
0 0
and (A + )50 � 50A = a b
c d
then [MT-AL] [301]
ekuk A =0
0 0
vkSj (A + )50 � 50A = a b
c d
rc
(A*) a + d = 2 (B*) a + b = 1 (C*) b + c = 0 (D) a + d = 1
Sol. A = 0
0 0
A2 = 0 0
0 0
= 0
A3 = 0
(A + )3 = A3 + 3A2 + 3A +
Similarly blh izdkj
(A + )50 = 50A + ......(1)
(A + )50 � 50A = 1 0
0 1
ans. A,B,C
a =1, b = 0, c = 0, d = 1
2. From a pack of 52 playing cards, all face cards are removed. From the remaining pack, cards are dealt one by one until an ace appear. Let P1 denote the probability that exactly 10 cards are dealt before the first ace
[Type text] Page 2
appear. If cards continue to dealt until a second ace appears. Let P2 denote the probability that exactly 20
cards are dealt before the second ace. If 130 P2 = mn
(m, n are co-prime natural numbers) then-
rk'k d h 52 iÙkksa dh ,d xM~Mh ls lHkh psgjs okys iÙks ¼face cards½ gVk;s tkrs gSA 'ks"k cph gqbZ xM~Mh ls ,d ,d
djds iÙks dks [khapk tkrk gS tc rd dh bDdk u vk tk;sA ekuk P1 ] bDdk vkus ls igys Bhd 10 iÙksa fudkyus dh
Further iqu% m + n < 2 m mn (m + n)! | (mn) and vkSj m � n < m < mn (m � n)! | (mn) 4. Let be an operation defined on points in Cartesian plane such that if 'P' and 'Q' are 2 points, then
PQ= R, where PQR is an equilateral triangle with P, Q, R being in anticlockwise sense, then
[MI-STAR] [303]
(A*) If PQ = R RP = Q
(B) If PQ = R PR = Q
(C) (PQ)S = P(QS)
(D*) If PQ = R, RQ = S and QS = T, then ratio of areas of PQR and QST is 1
ekuk dkfrZ; lery esa ifjHkkf"kr fcUnqvksa ij ,d lafØ;k bl izdkj gS fd P vkSj Q nks fcUnq ds fy, PQ= R, tgk¡
PQR leckgq f=kHkqt gS vkSj P, Q, R dks okekorZ fn'kk esa fy;k x;k gS] rc
Sol. Clearly if PQ = R, then as long as cyclic order is maintained by operating on any two we will get the third point, i.e. PQ = R RP = Q QR = P. Now for three points the operation need not be associative for example if PQ= S, then (C) will not hold true. In all the cases in (D) triangles have same side lengths hence their area are equal hence (A) and (D).
5. Let C1 and C2 be centres of two circles whose radii are 2 and 4 respectively. Also C1C2 = 10 and direct common tangents of these circles touch them at P,Q,R,S. Another circle of radius '' is drawn passing through P, Q, R, S, then [CR-CT] [301]
(A*) Mid-point of C1C2 is centre of the circle passing through P,Q,R,S.
(B) Centre of the circle passing through P,Q,R,S divides C1C2 in the ratio 1 : 2.
(C*) de ls de ,d c (0, 2) bl izdkj gS fd f '(c) = 0
(D*) f dk U;wure eku �2
gSA
Sol. f (x) = 0
cos tcos(x t)dt
....(1)
= 0
cos t·cos(x t)dt
f (x) = 0
cos t·cos(x t)dt
....(2)
(1) + (2) gives
2 f (x) = 0
cos t(2cos x·cos t)dt
f (x) = cos x 2
0
cos t dt
= 2 cos x2
2
0
cos t dt
f (x) = cos x2
Now verify vc lR;kfir.
9. Let Pn = n(3n)!(2n)!
(n = 1, 2, 3,.......) then n
n
P2lim
3 n is [LT-LG] [302]
ekuk Pn = n(3n)!(2n)!
(n = 1, 2, 3,.......) rc n
n
P2lim
3 n gS -
(A) 274e
(B*) 9e
1�cos x
2x 0
e �1lim
x
(C) e3
23 (D*)
92e
Sol. nPn
= 1/n
n
((2n 1)(2n 2).......(2n n))lim
n
[Type text] Page 8
nPn
= 1/n
2n 1 2n 2 2n 3 2n n.........
n n n n
logenP
n=
n
er 1
1 2n rlog
n n
logenP
n=
n
enr 1
1 rlim log 2
n n
dx)x2(lognP
loglim1
0
en
en
nen
Plim log
n=
27n
4e
nlim nP
n=
274e
10. If yx � xy = 1, then the value of dydx
at x = 1 is - JPT-1 [MD-GN] [303]
;fn yx � xy = 1 rc x = 1 ij dydx
dk eku gS -
(A) n(4e2) (B*) n2e
4
(C*) 2(1 � n 2) (D) 2(1 + n 2)
Sol. yx � xy = 1 [if x = 1 then y = 2] [;fn x = 1 rc y = 2]
Let ekuk P = yx
n P = x n y
1P
dPdx
= n y + xy
dydx
dPdx
= Px dy
nyy dx
dPdx
= yx x dyny
y dx
....... (1)
Let ekuk Q = xy
n Q = y n x
1Q
dQdx
=yx
+ n x . dydx
[Type text] Page 9
dQdx
= xy y dy
nx.x dx
........ (2)
Now vc yx � xy = 1 P � Q = 1
differentiate both sides w.r.t. x
x ds lkis{k vodyu djus ij
dPdx�
dQdx
= 0 yx x dyny
y dx
� xy
y dynx.
x dx
= 0 {from (1) and (2)} {(1) vkSj (2) ls}
when tc x = 1, y = 2 dydx
= 2(1 � n 2)
11. The curve, with the property that the projection of the ordinate on the normal is constant and has a length equal to 'a', may be (Where c is are abitrary constant) [TN-NR] [302]
oØ ftldh] dksfV dk vfHkyEc ij iz{ksi vpj gS rFkk a ds cjkcj yEckbZ gS] gks ldrk gS&
(tgk¡ c LoSPN vpj gS)
(A*) 2 2aln y a y x c (B) 2 2x a y c
(C) (y � a)2 = cx (D*) | y + 2 2y � a | = cex/a
Sol. Ordinate = PM. Let P (x, y) [T/S]
dksfV = PM. ekuk P (x, y)
Projection of ordinate on normal = PN
dksfV dk vfHkyEc ij iz{ksi = PN
PN = PM cos = a (given fn;k x;k gS)
2
ya
1 tan
y = 21a 1 (y )
2 2y ady
dx a
2 2
adydx
y a
2 2aln|y y a | x c
12. If z � axis be vertical, then the equation of the line of greatest slope through the point (2, �1, 0) on the plane 2x + 3y � 4z = 1 is [TD-PP] [304]
(� ) (� ), (�) (� ), ( � ) (� ) gS] rc fuEu esa ls lgh gS&
(A*) 24a + b + c = 0 (B*) b = 0 (C) a + b + c = 9 (D*) abc = 0
Sol. + + = 0
+ + = 3
= �2
Let ekuk y = (� ) (� )
y = 2 � (+ ) +
= 2 � (+ + ) + 2
y = 2 � 3 + 2
y = 3 � 3� 4
we know that 3 = � 3� 2
ge tkurs gS 3 = � 3� 2
y = � 6(+ 1) = �6
y 6
so equation is y3 + 9y2 � 216 = 0
lehdj.k y3 + 9y2 � 216 = 0 gSA
14. Sum of first n terms of the series 1 + 3 + 5 + 7 + ..... is 100. If 3 consecutive terms be removed from these n terms then sum of the remaining terms and sum of original terms are in the ratio 17 : 20. Which of the following is/are correct. [SS-AP] [303]
(A*) sum of the square of removed terms is 83
(B*) all 3 removed terms are prime
(C*) n = 10
(D) If , , are removed terms then coefficient of x in (x + )(x + )(x + ) is 81
[Type text] Page 12
Js.kh 1 + 3 + 5 + 7 + ..... ds izFke n inksa dk ;ksxQy 100 gSA ;fn 3 Øekxr inksa dks bu n la[;kvksa ls gVk;k tk,
(D) ;fn , , gVk, x, in gS rc (x + )(x + )(x + ) esa x dk xq.kkad 81 gSA
Sol. n2
[2 + (n � 1)2] = 100
n2 = 100 n = 10
sumof remainingterms
sumof original terms =
1720
= 85
100
so sum of 3 removed terms = 15
let these are a � d, a, a + d
a = 5
d = 2 terms are 3, 5, 7 hence A, B, C are correct
Hindi n2
[2 + (n � 1)2] = 100
n2 = 100 n = 10
'ks"k ink sadk ;ksxQy ewy ink sadk ;ksxQy
= 1720
= 85
100
blfy, gVk;h xbZ 3 la[;kvksa dk ;ksxQy gS = 15
ekuk ;s la[;k,¡ a � d, a, a + d
a = 5
d = 2 in 3, 5, 7 gSA vr% A, B, C lgh gSA
15. The range of values of m for which the line y = mx and the curve y =2
x
x 1enclose a region, lies in set
m ds ekuksa dk ifjlj ftlds fy, js[kk y = mx vkSj y =2
x
x 1 ,d {ks=k dks ifjc) djrs gS] ftl leqPp; esa fLFkr gS]
og gS& [SL-MS] [304]
(A*) (�1, 1) (B*) (0, 1) (C*) [0, 1] (D) (1, )
[Type text] Page 13
Sol. Solving gy djus ij
mx =x
x 1 x2 + 1 =
1m
or x = 0
x2 = 1m
� 1 > 0 for a region {ks=k ds fy,
m 1
m
< 0 m (0, 1)
Note: for m = 0 or m = 1 the line does not enclose a region. uksV: m = 0 ;k m = 1 ds fy, js[kk {ks=k dks ifjc) ugha djrh gSA
SECTION � 3 : Matching List Type (Only One Option Correct) [k.M � 3 : lqesyu lwph izd kj (dsoy ,d fodYi lgh)
This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A),(B),(C) and (D), out of which ONE is correct.
16. Match List I with List II and select the correct answer using the code given below the lists : lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s % [MTC-MS] [302] List-I List-II lwph- I lwph - II (P) If quadratic eqution (2 cos )x2 � (2 cosec)x + (cosec2
) = 0 1. 2 has only one solution for x, then number of values of
(0 2)are
(Q) The value of 2
2 2
0
x(sin (sinx) cos (cos x))dx
is equal to 2. 3
(R) If e is eccentricity of ellipse whose area is half the area of its auxiliary 3. 4
circle then 4e2 =
(S) If t0 is term independent of x in expansion of 9
2 1x �
x
and 1mt ,
2mt 4. 7
are coefficient of two middle terms then 1 20 m mt t t
12
feyku dhft, -
[Type text] Page 14
lwph- I lwph - II (P) ;fn f}?kkr lehdj.k (2 cos )x2 � (2 cosec)x + (cosec2
S.No. Subject Nature of Questions No. of Questions Marks Negative Total
1 to 15 MCQ 15 4 0 60
16 to 19 Match matrix listing 4 3 �1 12
20 to 34 MCQ 15 4 0 60
35 to 38 Match matrix listing 4 3 �1 12
39 to 53 MCQ 15 4 0 60
54 to 57 Match matrix listing 4 3 �1 12
57 216
Paper-2
TotalTotal
PAPER-2
SECTION-1 : (One or more option correct type) [k.M�1 : (,d ;k vf/kd lgh fodYi çdkj)
This section contains 15 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
bl [k.M esa 15 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d ;k vf/kd lgh gSA
MCQ._(15)
20. Two waves are simultaneously passing through a string, their equations are y1 = A1 sin k(x � vt) and y2 = A2 sin k (x � vt + x0) respectively, where k = 3.14 cm�1 and x0 = 3.0 cm, A1 = 6 mm, A2 = 5.2 mm, select the correct statements. [ST-SO](102)
(c) Amin = |A1 - A2| = |6 - 5.2| = 0.8 mm 21. A black body has a temperature of 'T' (in kelvin) and wavelength �� corresponding to maximum
energy density. When body cools, if the wavelength corresponding to the maximum energy density changes by 900% then choose the correct alternative(s) [HT-FA](104)
,d d f̀".kd k oLrq dk rki 'T' (dSfYou esa) rFkk vf/kdre~ Å tkZ ?kuRo ds laxr rjaxnS/;Z �� gSA tc oLrq BaMh
gksrh gSA ;fn vf/kdre~ Å tkZ ?kuRo ds laxr rjaxnS/;Z 900% ls ifjorfrZr gksrh gS rc lgh fodYiksa dk p;u
dhft,A
(A) Temperature of black body to which it is cooled is T9
(B*) Ratio of rate of emission i.e. final to initial will be 1 : 10000
(C*) The magnitude of difference in two temperature 9T10
(D) Ratio of rate of emission i.e. final to initial will be 1 : 6561
(A) d f̀".kd k oLrq dk rki ftl ij ;g BaMh gksrh gS] T9
(C*) magnitude of potential difference between B and D is 30 V.
B rFkk D ds e/; foHkokUrj dk ifjek.k 30V gSA
(D) magnitude of potential difference between B and C is 15 V.
B rFkk C ds e/; foHkokUrj dk ifjek.k 15V gSA
Ans. (bc)
Sol. 24 15 6
IR 1 2 1
A BV V 6 Ir
33
9 6R 4
R = 7
VBC = 15 � 3(2) = 9V
VBD = 30V
23. A block of mass 'm' rests on a fixed incline plane of inclination '' with horizontal. Assume friction is large enough to make the block stationary. Then choose correct alternative(s). [FR-MQ] (104)
m nzO;eku dk ,d CykWd {kSfrt ls dks.k ij >qds fLFkj urry ij fojke ij gSA ;g ekfu, fd ?k"kZ.k CykWd
24. A particle travels along the path y = a + bx + cx2 where a, b, c are positive constants. If v0 is the
speed of the particle and it is constant. Then choose the correct options, if at the given instant particle is at x = 0. [CM-VT](104)
,d d.k iFk y = a + bx + cx2 ds vuqfn'k xfr'khy gS tgk¡ a, b, c /kukRed fu;rkad gSaA ;fn v0 d.k dh pky
gS rFkk ;g fu;r gSA rc lgh fodYiksa dk p;u djks] ;fn fn;s x;s {k.k ij d.k x = 0 ij gS
(A*) radius of curvature is 2 3/2(1 b )
2c
oØrk f=kT;k 2 3/2(1 b )
2c
gksxh
(B*) magnitude of net acceleration is
20
3/22
2cv
1 b
dqy Roj.k dk ifjek.k
20
3/22
2cv
1 b gS
(C) radius of curvature is
2 20
3.22
2c v
1 b
oØrk f=kT;k
2 20
3.22
2c v
1 b gksxh
(D) magnitude of centriptal acceleration 2
02
v
1 b
vfHkdsfUnz; Roj.k dk ifjek.k 2
02
v
1 b gksxk
Sol. (ab)
R =
3/22
2 3/2
2
2
dy1
dx (1 b )2cd y
dx
r ta a a
ta
= 0. (speed constant) (pky fu;r gS)
| a
| = | ra
| = 20v
R =
20
3/22
2cv
1 b
25. In the circuit shown below the switch between A & B is closed at t = 0, then choose the correct options. (Consider circuit to be in steady state at t < 0 )
uhps n'kkZ;s x;s ifjiFk esa A o B ds e/; dqath t = 0 ij cUn dh tkrh gS, rc lgh fodYi@fodYiksa dk p;u
dhft,A (ekfu;s fd ifjiFk t < 0 ij LFkk;h voLFkk esa gSA)
B
R1
L
A E
R2
(A*) Current through R1 and R2 will not change just after the switch is closed
dqath cUn djus ds Bhd ckn R1 o R2 ls xqtjus okyh /kkjk ifjofrZr ugha gksrh gSA
(B) Current through R1 and R2 will change just after switch is closed
dqath cUn djus ds Bhd ckn R1 o R2 ls xqtjus okyh /kkjk ifjofrZr gksrh gSA
(C) Current through L will be different at both instants. i.e., just after switch is closed and after long time.
nksuksa {k.kksa ¼vFkkZr~ dqath cUn djus ds Bhd ckn ,oa cgqr yEcs le; i'pkr~½ ij L ls xqtjus okyh /kkjk
fHkUu&fHkUu gksxhA
(D*) Current through R2 will be same at t < 0 and t .
R2 ls xqtjus okyh /kkjk t < 0 ij rFkk t ij leku gksxhA
Sol. Since initially inductor L has current 1R
E flowing through it, so when switch is closed inductor will
give some induced emf, hence (A), (D)
pqafd izkjEHk esa izsjdRo L ls xqtjus okyh /kkjk 1R
E gS rFkk tc dqath dks cUn fd;k tkrk gS izsjdRo dqN izsfjr
fo-ok-cy nsrk gS vr% (A), (D)
26. Two spring of spring constant K and 4K are attached with a block of mass m and other end of springs are free, system is placed between two rigid wall W1 and W2. Springs are in its natural
length at instant shown, at t = 0 a sharp impulse is give to the block towards wall W1 then
Sol. Two springs are not compressed or extended together and compression of both springs are not same. So block not perform SHM, it perform oscillatory motion and time period is given by
nksuksa fLiazx ,d lkFk lEihfMr ;k izlkfjr ugha gksrh gS rFkk nksuksa fLiazx esa lEihM+u leku ugha gSA vr% ljy
28. An infinitely long straight wire carrying a current 1 is partially surrounded by ABCD loop as shown in figure, arc AD and BC have circular shape and the infinite wire passes through their centre C1C2. The loop has a length L, radius R and carries a current 2. The axis of the loop coincides with the wire, ABCD plane and infinite length wire are coplanar. Then
29. A charged particle (+q) is moving simple harmonically on the x-axis with its mean position at origin. Amplitude of the particle is A and its angular frequency is .
,d vkosf'kr d.k (+q), x-v{k ij ljy vkorZ xfr djrk gS] ftldh ek/; fLFkfr eqy fcUnq ij gSA d.k dk
vk;ke A rFkk bldh dks.kh; vkof̀Ùk gSA [EM-FQ](104)
y
x (A, 0) (�A, 0) (0, 0)
(A) The magnitude of magnetic field at (2A, 0) will change periodically with period 2/.
(B*) The maximum magnitude of the magnetic field at (0, A) is Aq
40
(C*) The magnetic field at (A, A) at the moment the particle passes through
0,
2A
, will be
A55
q3 0
(D*) The magnitudes of magnetic field at (0, A) and (0, �A) will be same at any time.
(A) (2A, 0) ij pqEcdh; {ks=k dk ifjek.k vkorZdky 2/ ls vkorhZ :i ls ifjofrZr gksxkA
(B*) (0, A) ij pqEcdh; {ks=k dk vf/kdre ifjek.k Aq
40
gSA
(C*) d.k tc
0,
2A ls xqtjrk gS ml {k.k (A, A) ij pqEcdh; {ks=k
A55
q3 0
gksxkA
(D*) (0, A) rFkk (0, �A) ij pqEcdh; {ks=k dk ifjek.k fdlh Hkh le; ij leku gksxkA
Sol. (a) For (2A, 0), = 0° or 180°
B = 0 permanent zero LFkk;h :i ls 'kwU;
(b) Magnetic field will be max at (0, A) when the particle passes through (0, 0)
(0, A) ij pqEcdh; {ks=k vf/kdre gksxk tc d.k (0, 0) ls xqtjrk gSA
30. The radii of a spherical capacitor are equal to a and b (b > a). The space between them is filled with a dielectric of dielectric constant K and resistivity . At t = 0, the inner electrode is given a charge q0. Choose the correct options : [CP-EQ](103) [Capacitance]
31. A small object moves counter clockwise along the circular path whose centre is at origin as shown in figure. As it moves along the path, its acceleration vector continuously points towards point S. Then the object [CH-HZ](104)
(A*) B ls gksdj A ls C rd xfr ds nkSjku pky c<+rh gSA
(B) B ls gksdj A ls C rd xfr ds nkSjku pky ?kVrh gSA
(C*) D ls gksdj C ls A rd xfr ds nkSjku pky ?kVrh gSA
(D) D ls gksdj C ls A rd xfr ds nkSjku pky c<+rh gSA
Sol. (Moderate) As the object moves from A to C via B the angle between acceleration vector and velocity vector decreases from 90° and then increases back to 90°. Since the angle between
velocity and acceleration is acute, the object speeds up. As the object moves from C to A via D the angle between acceleration vector and velocity vector
increases from 90° and then decreases back to 90°. Since the angle between velocity and
acceleration is obtuse, the object slows down.
tSlk fd oLrq B ls gksdj A ls C xfr djrh gSA Roj.k lfn'k ,oa osx lfn'k ds e/; dks.k 90° ls ?kV+rk gS ,oa
,d jsfM;ks/kehZ {k; vfHkfØ;k esa % [MADE: CSS 2012-13] [NP-DL](102)
A 2
B 9
C
Select correct alternative/s at the instant the number of the particles of B is maximum :
(A*) Activity of A is equal to activity of B (B*) No of atoms of A is 4.5 times of B
(C) Activity of A is more then activity of B (D) Activity of A is minimum
lgh fodYi@fodYiksa dk p;u ml {k.k dhft, tc B ds d.kksa dh la[;k vf/kdre gS %
(A*) A dh lfØ;rk B dh lfØ;rk ds rqY; gSA (B*) A ds ijek.kqvksa dh la[;k B dh 4.5 xquk gSA
(C) A dh lfØ;rk B dh lfØ;rk ls vf/kd gSA (D) A dh lfØ;rk U;wure gSA
Sol. NA 2 = NB 9
NA = 4.5 NB
33. In series LCR circuit voltage drop across resistance is 8 V and voltage across inductor to is 6 V across capacitor is 12 volt. Then select incorrect alternative/s : [AC-RC](102)
Js.kh LCR ifjiFk esa izfrjks/k ds fljksa ij foHko iru 8 V gS rFkk izsjd dq.Myh ds fljks ij oksYVst 6 V gS rFkk
la/kkfj=k ds fljks ij 12 V gSA rc vlR; fodYi@fodYiksa dk p;u dhft,A
(A*) Voltage of the source will be leading current in the circuit
ifjiFk esa L=kksr oksYVst /kkjk ls vkxs ¼leading½ gksxkA
(B*) Voltage drop across each element will be less than the applied voltage
izR;sd vo;o ds fljks ij foHko iru vkjksfir oksYVst ls de gksxkA
This section contains 4 multiple choice questions. Each questions has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
bl [k.M esa 4 cgqfod Yi ç'u gSaA çR;sd ç'u esa lqesyu lwph gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C)
vkSj (D) gSa] ftuesa ls dsoy ,d lgh gSA
Match List_(4) 35. Type of ideal gas and process followed by it is given in list-I, and in list-II related known
parameters (P,V,T) or quantities or nature is given. Match list-Iwith list-IIand select the correct answer using the codes given below the lists . (where R is universal gas constant) [TH-AD](102)
lwph-I esa vkn'kZ xSl dk izdkj rFkk blds }kjk vuqlfjr izØe fn;k x;k gS rFkk lwph-II eas lEcfU/kr izkpy
36. Three metallic bars 1, 2, 3 are arranged as shown is figure, with number of free charge carriers in ratio N1 : N2 : N3 = 1 : 3 : 2; resisitivity ratio 1 2 3: : 2 : 1: 3; lengths in ratio
1 2 3: : 2 : 2 : 3 for 1, 2, and 3 bars respectively (radius of cross-section shown in figure),
carry current i as shown. Match list-with list-and select the correct answer using the codes given below the lists : [CE-DF](103)
37. The figures in list- show some charge and current distribution with a charged particle projected in some specific direction list- gives certain conditions which may exist in the subsequent motion of
38. An object O (real) is placed at focus of an equi-convex lens as shown in figure. The refractive
index of material of lens is = 1.5 and the radius of curvature of either surface of lens is R. The lens is surrounded by air. In each statement of list-I some changes are made to situation given above and information regarding final image formed as a result is given in list-II. The distance between lens and object is unchanged in all statements of list-I. Match list-with list-and select the correct answer using the codes given below the lists : [M.Bank_GO_12.14]
[GO-LE](102)
,d fcEc O (okLrfod ) fp=k ds vuqlkj ,d lemÙky (equi-convex) ysUl ds Qksdl ij fLFkr gSA ysUl ds
Codes : P Q R S (A) 1 2 3 4 (B*) 4 3 2 3 (C) 4 3 2 1 (D) 3 1 4 2 Ans. (P) � 1,3,4 ; (Q) 2, 3 ; (R) 2, 3 ; (S) 2, 3
Sol. Initially the image is formed at infinity.
(P) As increases the focal length decreases. Hence the object is at a distance larger than focal length. Therefore final image is real. Also final image becomes smaller in size in comparision to size of image before the change was made.
(Q) If the radius of curvature is doubled, the focal length increases. Hence the object is at a distance lesser than focal length. Therefore final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.
(R) Due to insertion of slab the effective object for lens shifts right wards. Hence final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.
(S) The object comes to centre of curvature of right spherical surface as a result. Hence the final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.
izkjEHk esa izfrfcEc vuUr ij curk gSA
(P) tc dks c<+k;k tkrk gS] rks Qksdl nwjh ?kVrh gSA blfy;s fcEc Qksdl nwjh ls vf/kd nwjh ij gksxkA
Course : JP, JF, JR JPT-1 (JEE ADVANCE) Test Date : 10-05-2015
Test Type : JEE ADVANCED (ELPD)
SYLLABUS : FULL SYLLABUS
Test Pattern :
S.No. Subject Nature of Questions No. of Questions Marks Negative Total
1 to 12 SCQ 12 3 �1 36
13 to 20 Comprehenstion (4 x 2Ques) 8 3 �1 24
21 to 25 Integer Type Questions (Two Digits Answer) 5 3 0 15
26 to 37 SCQ 12 3 0 36
38 to 45 Comprehenstion (4 x 2Ques) 8 3 �1 24
46 to 50 Integer Type Questions (Two Digits Answer) 5 3 0 15
51 to 62 SCQ 12 3 0 36
63 to 70 Comprehenstion (4 x 2Ques) 8 3 �1 24
71 to 75 Integer Type Questions (Two Digits Answer) 5 3 0 15
75 225
S.No. Subject Nature of Questions No. of Questions Marks Negative Total
1 to 15 MCQ 15 4 0 60
16 to 19 Match matrix listing 4 3 �1 12
20 to 34 MCQ 15 4 0 60
35 to 38 Match matrix listing 4 3 �1 12
39 to 53 MCQ 15 4 0 60
54 to 57 Match matrix listing 4 3 �1 12
57 216
Paper-1
Paper-2
Total Total
TotalTotal
Maths
Physics
Chemistry
Physical Inorganic Chemistry Paper-1 Organic Chemistry Paper-1 SCQ (8) SCQ (4) Comp.(4 x 2Q.) (2) Comp.(4 x 2Q.) (2) Integer (Double digit) (3) Integer (Double digit) (2) Physical Inorganic Chemistry Paper-2 Organic Chemistry Paper-2 MCQ (10) MCQ (5) Match Listing type (2) Match Listing type (2)
Page # 2
Remarks (if any):
Page # 3
JEE (ADVANCED) CHEMISTRY PAPER SKELETON
Faculty Name : Test Name : JR (JPT-1) Faculty preparing the TEST PAPER should fill it according to paper pattern and submit it with finalisation of paper at SMD.
PAPER-2 MCQ (10) 39. NaCl (aq.) solution is taken at 0ºC and 1 atm. Now some ice is added in it keeping temperature and
pressure constant then : (SCP-EBP_207(P)) (A*) Vapour pressure of solution will increase (B) Osmotic pressure of solution will increase (C*) Whole of the added ice will melt (D) Some amount of ice(s) will exist in solution NaCl (tyh;) foy;u dks 0ºC rFkk 1 atm ij fy;k x;k gS vc rki rFkk nkc dks fu;r j[krs gq, blesa dqN cQZ
feykrs gS] rc & (A*) foy;u dk ok"inkc c<+sxkA (B) foy;u dk ijklj.k nkc c<+sxkA (C*) feyk;h xbZ lEiw.kZ cQZ fi?ky tk,xhA (D) cQZ dh dqN ek=kk foy;u esa fo|eku gksxhA Sol. Since freezing point of solution is lower than pure water so whole of the added ice will melt. Due to this
solution will be diluted and vapour pressure will increase. pwafd foy;u dk fgekad 'kq) ty ls de gS blfy, feyk;h xbZ lEiw.kZ cQZ fi?ky tk,xhA ftlds dkj.k foy;u ruq
gksxk rFkk ok"inkc c<+sxkA 40. Cold NaCl aqueous solution is electrolysed with vigorous stirring then correct option(s) is/are : (PBC-XVII_206(I)) (A*) Process will involve disproportionation reaction. (B*) Final product has bleaching action. (C) Same product is formed at high temperature (D*) Gas formed at cathode can not be dried by conc. H2SO4. BaMs NaCl ds tyh; foy;u dks rsth ls foyksfMr (stirring) gq, bls fo|qr vi?kfVr djrs gS tc lgh fodYi@fodYiksa
dk p;u dhft,A (A*) izØe esa fo"kekuqikrhdj.k vfHkfØ;k gksxh (B*) vfUre mRikn fojatd fØ;k djrk gSA (C) mPp rki ij leku mRikn curk gSA (D*) dSFkksM ij fufeZr xSl dks lkUnz H2SO4 }kjk 'kq"d ugha fd;k tk ldrk gSA Sol. At Cathode dSFkksM ij 2H2O + 2e� H2 + 2OH� At Anode ,uksM ij 2Cl� Cl2 + 2e�
Cl2 + OH� Cold ( )
B.Mk Cl� + OCl�
41. Colemanite 2 3Na CO x HCl
yMonoprotic
acid
z Mg
wan element
Identify incorrect statement. (PBC-XIII_206(I)) (A) Oxidation number of boron in colemanite is +3 (B*) z when heated with Co gives blue colour in oxidizing flame and red in reducing flame. (C*) Almost all compounds of w are ionic. (D*) Decahydrated x has 4 O�H bonds.
(B*) z dks tc Co ds lkFk xeZ djrs gS] rks ;g vkWDlhdkjh Tokyk esa uhyk jax rFkk vipk;ddkjh Tokyk esa yky jax nsrk gSA
(C*) w ds lHkh ;kSfxd izk;% vk;fud gksrs gSA (D*) Msdkty;ksftr x 4 O�H ca/k j[krk gSA
Sol. Ca2B6O11 + Na2CO3 2 4 7(X)
Na B O HCl 3 3
(Y)H BO
2 3(Z)
B O Mg
(W)B
Decahydrated Borax is Na2[B4O5(OH)4].8H2O Msdkty;ksftr cksjsDl Na2[B4O5(OH)4].8H2O gSA So it has 20 O�H bonds. vr% ;g 20 O�H ca/k j[krk gSA 42. Which of following represents equilibrium condition? (CEQ-HEE_206(P)) (A*) 10 g ice and 5 g H2O () at 0ºC & 1 atm (B*) 5 g ice and 5 g H2O () at 0ºC & 1 atm (C*) 10 g H2O () and 5 g H2O (v) at 100ºC & 1 atm (D*) 5 g H2O () and 5 g H2O (v) at 100ºC & 1 atm fuEu esa ls dkSu lkE; ifjfLFkfr iznf'kZr djrs gSa\ (A*) 0ºC rFkk 1 atm ij 10 g cQZ rFkk 5 g H2O () (B*) 0ºC rFkk 1 atm ij 5 g cQZ rFkk 5 g H2O ()
(C*) 100ºC rFkk 1 atm ij 10 g H2O () rFkk 5 g H2O (v)
(D*) 100ºC rFkk 1 atm ij 5 g H2O () rFkk 5 g H2O (v) Sol. P and T condition decides equilibrium for such equilibria (not amount). nkc rFkk rki ifjfLFkfr bl izdkj ds lkE;ksa ds fy, lkE; dks fuf'pr djrh gSA ¼ek=kk ugha½ 43. Salt AB undergoes anionic hydrolysis and its 0.1 M solution has pOH as 5 then. Identify correct. (IEQ-AS_206(P)) (A*) Kh is equal to Kb of B� (B*) pKa of HB is 5 (C*) h is 0.01% (D*) pH of 0.1 M HB is 3 yo.k AB _ .kk;fud ty vi?kVu nsrk gS rFkk bldk 0.1 M foy;u 5 pOH j[krk gS] rc lgh dFku igpkfu;sA (A*) Kh , B
� ds Kb ds cjkcj gksrk gSA (B*) HB dk pKa 5 gSA (C*) h 0.01% gSA (D*) 0.1 M HB dh pH 3 gSA Sol. B� + H2O HB + OH� c(1�h) ch ch c = 0.1 M ch = 10�5 h = 10�4 Kh = ch2 = 10�9 Ka(HB) = 10�5 Kb of B� = 10�9
pH = 12
(5 � (�1)) = 3
44. A solid A+B� has r 2 2R 2
then correct statement is : (SST-RR_206(P))
(A*) B� touches total 18 ions (B*) A+ touches 6 ions (C*) Each body diagonal passes through the centre of 2 cation and 1 anion (D*) Each body diagonal passes through the centre of 2 anion and 1 cation
Sol. This radius ratio suggests exact fitting in rock salt type structure. ;g f=kT;k vuqikr jkWd lkYV izdkj dh lajpuk esa iw.kZr% O;ofLFkr gksrk gSA 45. Which is/are correct about SO2Cl2 ? (PBC-XV_206(I))
(A*) OSO > OSCl > ClSCl bond angle (B*) Number of p� p bonds < Number of p�d bonds (C*) 1 mol of this reacts with 0.1 mol P4 (D*) Its aqueous solution does not give pink colour with phenolphthaleine SO2Cl2 ds lUnHkZ esa dkSulk@dkSuls dFku lgh gS@gSa\
46. Compound of Na Na, A + colourless gas B. (SBC-CAM_206(I))
(x) Identify true statement : (A*) Cl2 disproportionates in aqueous solution of A. (B*) B does not show allotropy (C*) X can give red colouration if treated with NH2CSNH2 + CH3COOH then followed by FeCl3 (D) B is not combustible but combustion supporter.
mipkfjr fd;k t krk gSA (D) B nguh; ugh gS] ysfdu ;g ngu lgk;d gksrk gSA Ans. x NaNO2 or ;k NaNO3
A Na2O B N2
47. CaCO3(s) CaO(s) + CO2(g) (CEQ-SEQ_207(P))
CO2(g) CO(g) + 12
O2 (g)
For above simultaneous equilibrium if CO2 is added from out side at equilibrium then :
mijksDr ,d lkFk gksus okys lkE;ks ds fy, ;fn CO2 dks lkE; ij vyx ls ¼ckgj ls½ feyk;k tkrk gS rc &
(A) 2COP will increase (B)
2COP will decrease
(C*) No shift in 2nd equilibrium (D*) Backward shift in 1st equilibrium (A)
2COP c<sxkA (B) 2COP ?kVsxkA
(C*) 2nd lkE; esa foLFkkiu ugha gksxkA (D*) 1st lkE; esa i'p fn'kk dh vksj foLFkkiu gksxkA Sol. Second equilibrium will not be affected by CO2 addition only first will shift backward.
Page # 7
f}rh; lkE; CO2 feykus ls izHkkfor ugha gksxk] dsoy izFke lkE; gh i'p fn'kk d h vksj foLFkkfir gksxkA 48. An aqueous solution of ZnCl2 with conc. H2SO4 is electrolysed using zinc electrodes at anode and cathode
then which of following options are correct ? (ECH-AOE_205(P)) (A) Cl2 gas can evolve at anode due to over voltage conditions (B*) H2 gas evolve at cathode so pH of solution get increases (C) Zinc will oxidise at anode but conc. of Zn+2 of solution remain constant. (D*) conc. of Zn+2 in electrolyte increases but conc. of anion remain same lkUnz H2SO4 ds lkFk ZnCl2 ds tyh; foy;u dks ,uksM rFkk dSFkksM ij ftad bysDVªkWM dk iz;ksx djds fo|qr
vi?kfVr fd;k tkrk gS rc fuEu esa ls d kSuls fodYi lgh gSa \ (A) Cl2 xSl vf/kd oksYVst ifjfLFkfr;ks ds dkj.k fu"dkflr gks ldrh gSA (B*) H2 xSl dSFkksM ij fu"dkflr gksrh gS blfy, foy;u dh pH c<+rh gSA (C) ftad ,uksM ij vkWDlhd r̀ gksxh ysfdu foy;u ds Zn+2 dh lkUnzrk fu;r jgrh gSA (D*) fo|qr vi?kV; esa Zn+2 dh lkUnzrk c<rh gSA ysfdu _ .kk;u dh lkUnzrk leku jgrh gSA Sol. Electrolyte ZnCl2 and conc. H2SO4 electrode Zn rod
Anode reaction Zn Zn+2 + 2e�
Cathode reaction 2H+ + 2e� H2 So conc. of [Zn+2] , conc. of Cl� and SO4
Which amongs the following is/are correct for above sequence of reactions ? (A*) Oxidative ozonolysis product of P, R & S will be achiral (B*) R & S are diastereomers (C*) Product Q & R are formed by syn addition & S is formed by anti addition (D) P must have cis configuration
H2 + Pd + BaSO4
Na / liq. NH3
fy.Mykj dk mRizsjd
;kSfxd (Q) C8H18 H2/Pt
;kSfxd (P) C8H12
(izdkf'kd lfØ;)
(izdkf'kd vfØ;)
;kSfxd (R) C8H14 (izdkf'kd lfØ;)
;kSfxd (S) C8H14 (izdkf'kd vfØ;)
mijksDr vfHkfØ;k vuqØe ds fy, lgh dFku@dFkuksa dk p;u dhft,& (HYC-RAA(O)(M)_203) (A*) P, R o S dk vkWDlhdkjh vkstksuhvi?kVu mRikn vfdjsy gksxkA (B*) R o S foofje leko;oh gSA (C*) mRikn Q o R flu ;ksx ls fufeZr gksrk gS o mRikn S ,UVh ;ksx ls fufeZr gksrk gSA (D) ;kSfxd P dk lei{k foU;kl gksuk pkfg,A
Page # 8
Sol.
H3C�CH=CH�CH�CC�CH3
| CH3
(P)
H3C�CH2�CH2�CH�CH2�CH2�CH3
| CH3
CH3
(Trans) | (Cis) CH3�CH=CH�CH�CH=CH�CH3
CH3
| CH3�CH=CH�OH�CH=CH�CH3
(Trans) (Trans)
Trans
Sol.
H3C�CH=CH�CH�CC�CH3
| CH3
(P)
H3C�CH2�CH2�CH�CH2�CH2�CH3
| CH3
CH3
(foi{k) | (lei{k) CH3�CH=CH�CH�CH=CH�CH3
CH3
| CH3�CH=CH�OH�CH=CH�CH3
(foi{k) (foi{k)
foi{k
50. Which statements are correct in the following ? (AC-AA(O)(E)_203) (A*) Cyclopentadiene reacts with NaNH2 and releases NH3 gas however cyclopropene does not. (B*) Cyclopropenone has a greater dipole moment than acetone (C*) Cyclobutadiene does not exist at room temperature but cyclobutene exists at room temperature (D*) Cycloheptatrienyl bromide ionizes faster however cyclopentadienyl bromide does not fuEu esa ls dkSuls dFku lgh gS\ (A*) lkbDyksisUVkMkbbZu NaNH2 ds lkFk fØ;k djds NH3 eqDr djrh gS tcfd lkbDyksizksihu ughaA (B*) lkbDyksizksiukWu dk f}/kzqo vk?kw.kZ ,lhVksu dh rqyuk esa vf/kd gksrk gSA (C*) lkbDyksC;wVkMkbbZu dejs ds rki ij ugha ik;k tkrk gS ysfdu lkbDyksC;wfVu dejs ds rki ij ik;k tkrk gSA (D*) lkbDyksgsIVkVªkbZbukbZy czksekbM rsth ls vk;fur gks tkrk gS ;|fi lkbDyksisUVkMkbZukbZy czksekbM ugha gSA
Sol. (A)
2NaNH
+ NH3
Aromatic
,
2NaNH
+ NH3
Antiaromatic
(B)
O
is aromatic & highly polar.
(C) is antiaromatic & is nonaromatic.
(D)
Br
Br
Aromatic,
Br
�Br
Antiaromatic
Sol. (A)
2NaNH
+ NH3
,sjkseSfVd
,
2NaNH
+ NH3
,UVh,sjkseSfVd
(B)
O
,jkseSfVd o mPp /kzqfo; gSA
(C) ,UVh,sjkseSfVd gS o ukWu,sjkseSfVd gS
Page # 9
(D)
Br
Br
,sjkseSfVd ,
Br
�Br
,UVh,sjkseSfVd
51. Optically active amines having molecular formula C5H13N on reaction with NaNO2 + HCl produces tertiary
optically inactive alcohol. Find out structures of amines. v.kqlw=k C5H13N ;qDr izdkf'kd lfØ; ,ehu dh NaNO2 + HCl ds lkFk vfHkfØ;k djkus ij rr̀h;d izdkf'kd vfØ;
Sol. (1) Needs primary amine for preparation of alcohol with HNO2. (NaNO2 + HCl) (2) Only (A) & (C) are optically active which after reaction with NaNO2 + HCl will give optically inactive
nsrh gSA 52. Which of the following are reducing sugar ? (BP-CBH(O)(E)_203) fuEu esa ls dkSulh vip;ukRed 'kdZjk gS\
(A)
OH
O OH
HO
OH
O HO
O
OH OH
HO
(B*)
OH
O OH
HO
OH O
OH O
OH
OH
OH
(C*)
(CHOH3)3 OH
HO O
(D*)
OH
O OH
HO
OH
O OH
O
OH
OH
OH
Sol. -Hydroxy carbonyl groups & sugars having hemiacetal group will be reducing sugar. -gkbMªksDlh dkcksZfuy lewg o 'kdZjk gsfe,sflVsy lewg j[krh gS tks vip;ukRed 'kdZjk gksxhA 53. Which of the following reactions give same product ? fuEu esa ls dkSulh vfHkfØ;k leku mRikn nsrh gSA (AC-EAS(O)(M)_204)
(A*)
NO2
H3C
Sn
HCl 2NaNO
HCl 3 4H PO
(B*)
3
CO HCl
AlCl
4LiAlH
TsCl
4NaBH
(C)
3
3
O||
CH C Cl
AlCl
4KMnO H
4LiAlH
(D*)
Cl
AlCl3
4KMnO H
Red P
H
Sol. (A*)
NO2
H3C
Sn
HCl
NH2
H3C
2NaNO
HCl
N2Cl
H3C
3 4H PO
H3C
Page # 10
(B*)
3
CO HCl
AlCl
CHO
4LiAlH
CH2�OH TsCl
CH2OTs
4NaBH
CH3
(C)
3
3
CH COCl
AlCl
COCH3 4KMnO
COOH
4LiAlH
CH2OH
(D*)
Cl
AlCl3
4KMnO H
COOH Red P
H
P
H
yky CH3
Match Listing type (MTC)(2) 54. Match the following : Column � I Column � II (PBC(INO)) (P) SiO2 (1) Reacts with HF
(Q) [Co(CO)4]� (2) Pseudo halide
(R) I� (3) Gives compound with Cu2+ via redox reaction (S) N2 (4) Inert towards reaction LrEHk � I LrEHk � II (P) SiO2 (1) HF ds lkFk fØ ;k
Ans. (A) � p, r ; (B) � p, s ; (C) � p ; (D) � q, s Sol. (A) is cannizzaro reaction (B) is Aldol condensation (C) is Nucleophilic addition (D) is cleaisen condensation Sol. (A) dSfutkjks vfHkfØ;k gSA (B) ,YMksy la?kuu gSa (C) ukfHkdLusgh ;ksxkRed vfHkfØ;k gSA (D) dSylu la?kuu gSA 57. Match the reaction in List-I with appropriate option in List-II. (CAD-AEA(O)(M)_204) (HYC-RAA(O)(M)_204) List-I List-II
(P) HBr (1) Substitution reaction
(Q) (2) Addition reaction
(R) (3) Coupling reaction
(S) (4) Racemic mixture
dkWye-I esa nh xbZ vfHkfØ;kvksa ds fy, dkWye-II ls mfpr feyku dhft;A lwph -I lwph-II
(P)
HBr (1) izfrLFkkiu vfHkfØ;k
(Q) (2) ;ksxkRed vfHkfØ;k
(R) (3) ;qXeu vfHkfØ;k
(S) (4) jslsfed feJ.k (Racemic mixture)
Code dwV : (P) (Q) (R) (S)
Page # 13
(A) 1,3 2,4 1,4 2,4 (B*) 2,4 1,3 1,4 2,4 (C) 2,4 1,3 2,4 1,4 (D) 1,4 2,4 1,3 2,4 Ans. (A) � q, s; (B) � p, r; (C) � p, s; (D) � q, s Sol. (A) The reactant undergoes electrophilic addition having a carbocation intermediate forming a racemic-
mixture. (B) The reactant undergoes an electrophilic substitution which is a coupling reaction without any
carbocation intermediate. (C) The reactant undergoes an electrophilic substitution reaction having a carbocation rearrangement
forming a racemic mixture. (D) The reactant undergoes nucleophilic addition forming a racemic mixture without any carbocation