MATHEMATICS MATHEMATICS Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected]Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029 P2JPT2(ADV.)170515C0-1 TEST PATTERN AIOT (JEE ADVANCE) - 2 TEST DATE : 03-05-2015 BATCH : S.No. Subject Nature of Questions No. of Questions Marks Negative Total 1 to 8 MCQ 8 3 0 24 9 to 14 Comprehension (3 Comp. x 2 Q.) 6 3 1 18 15 to 20 Integer Type Questions (Double Digits Answer) 6 3 0 18 21 to 28 MCQ 8 3 0 24 29 to 34 Comprehension (3 Comp. x 2 Q.) 6 3 1 18 35 to 40 Integer Type Questions (Double Digits Answer) 6 3 0 18 41 to 48 MCQ 8 3 0 24 49 to 54 Comprehension (3 Comp. x 2 Q.) 6 3 1 18 55 to 60 Integer Type Questions (Double Digits Answer) 6 3 0 18 60 180 Total Total Maths Physics Chemistry Paper-2 (AIOT-2 JEE ADVANCE) MCQ 8 1. Let a = sin 1 (sin3) + sin 1 (sin4) + sin 1 (sin5), [FN-IN] [307] f(x) = | x | x 2 e , domain of f(x) be [a, ), range of f(x) be [b, ), g(x) = (4cos 4 x 2cos2x 2 1 cos4 x x 7 ) 1/7 , domain and range of g(x) is set of real numbers. Which of the following are CORRECT ? (A*) a = 2 (B*) b + a = 1 (C*) f(gog(b)) =e 2 (D)both f(x), g(x) are non-invertible functions ekuk a = sin 1 (sin3) + sin 1 (sin4) + sin 1 (sin5), f(x) = | x | x 2 e , f(x) d k izkUr [a, ) gS] f(x) d k ifjlj [b, ), g(x) = (4cos 4 x 2cos2x 2 1 cos4 x x 7 ) 1/7 , g(x) d k izkUr vkSj ifjl j] okLrfod la[;kvksad k l eqPp; gSfuEu esal sd kSul k lgh gS ? (A*) a = 2 (B*) b + a = 1 (C*) f(gog(b)) =e 2 (D) f(x), g(x) nksuksaizfry kse Q y u ughagSA
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
MA
TH
EM
AT
ICS
MATHEMATICS
Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Sol. a = ( � 3) + ( � 4) + (5 � 2)=�2 f(�2) = f(2) f(x) is many one non invertible f(�2) = f(2) f(x) cgq,Sd h gSA izfry kseh; ugha gSA
3
4 4 5
Let ekuk t = x2+ |x| t )
X
t
f(x) [1, ) b = 1 a + b = �1
g(x) =((1 + cos 2x)2 �2cosx �
21
(2cos22x�1)�x7)1/7
=(1 + 2cos 2x+cos22x �2cos2x�cos22x + 21
� x7)1/7
g(x) =7/1
7x�23
gog(x) =7/1
7))x(g(�23
=7/1
7x�23
�23
= (x7)1/7
= x f(gog(b)) = f(b) = e1+1 = e2 2. Following usual notation, if in a [ST-HA] [303] triangle ABC, r1 + r2 = 3R, r2 + r3 = 2R, then smallest angle is 10nº, n N. The equation (x � 1) (x � 2)(x � 3) = 24 has real root and imaginary root ± iR.
Which of the following are CORRECT ? ABC esa lkekU; l ad srkuqlkj r1 + r2 = 3R, r2 + r3 = 2R, rc lcls NksVk d ks.k 10nº, n N. lehd j.k (x � 1) (x � 2)(x � 3) = 24 d s okLrfod ewy vkSj d kYifud ewy ± iR. fuEu
esa l s d kSulk lgh gS ? (A*) n = 3 (B*) n + = 8 (C*) 2 +2 = 6 (D*) + 2= 2n
�1 intersects the curve x2 � y2 = a2, z = 0 if a is equal to
js[kk x � 2
3=
y �1
2 =
z �1
�1 oØ x2 � y2 = a2, z = 0 d ks izfrPNsn d jrh gS] ;fn a cjkcj gS&
(A*) 4 (B) 5 (C*) �4 (D) � 5 Sol. For the point where the line intersects the curve, we have z = 0, so Revi. PT-3-29-4-2015 fcUnq d s fy ,] js[kk oØ d s fy , izfrPN sn d jrh gSA z = 0, blfy , [TD-MS] [303]
x � 2
3=
y �1
2=
0 �1
�1 x = 5 and vkSj y = 3
Put these values in x2 � y2 = a2, we get a2 = 16 a = ±4 x2 � y2 = a2 esa bu ekuksa d ks j[kus ij, a2 = 16 a = ±4
4. Consider the graph of the function f(x) =x 3
nx 1e
then which of the following is correct (A) Range of the function is (1, ) [FN-DN] [302]
(B*) f(x) has no zeroes (C*) Graph lies completely above the x-axis (D*) Domain of f is (� , �3) (�1, )
ekukfd Q y u f(x) =x 3
nx 1e
d k vkjs[k d s fy , rc fuEu esa ls d kSulk lgh gS& (A) Q y u d k ifjlj (1, ) gSA
(B*) f(x) 'kwU; ugha gSA (C*) vkjs[k iw.kZr;k x-v{k ls Å ij gSA
(D*) f d k izkUr (� , �3) (�1, ) gSA
Sol. y = 1x3x
> 0 x < � 3 or ;k x > � 1
or ;k x �3 x x �1 x 1 (0, 1) (1, )
5. If
0i2
2
0jji )1a()1�a(
a
a.a
1 where i j and a >1 then possible values of may be
;fn
0i2
2
0jji )1a()1�a(
a
a.a
1 t gk¡ i j vkSj a > 1 rc d s laHko eku gks ld rs gS&
(A) 1 (B) 2 (C*) 3 (D*) 4
Sol. When no restriction on i and j [SS-GP] [303] t c i vkSj j ij d ksbZ izfrcU/k ugha gSA
6. A manufacturer of airplane parts makes a certain engine that has a probabil ity p of
failing on any given f light. There are two planes fitted with this type of engine. One plane has 3 such engines and other planes has 5. A plane crashes if more than half the engines fit ted in it fail. If the two plane models have the same probabil ity of crashing then the value of p can be [PR-BT] [301]
,d gokbZ t gkt cukus d h d Eiuh] d qN bat u cukrh gS] mld s fd lh nh xbZ mMku d s [kjkc gksus d h izkf;d rk p gSA bl izd kj d s bat uksa ls nks gokbZ t gkt cuk, t krs gSA ,d gokbZ t gkt esa bl izd kj d s 3 bat u gS vkSj nwljs gokbZ t gkt esa 5 bat u gSA ,d gokbZ t gkt d h nq?kZVuk gksrh gSA ;fn y xk, x, vk/ks ls T;knk bat u [kjkc gksrs gSA nks gokbZ t gkt d h nq?kZVuk ?kfVr gksus d h izkf;d rk leku gS] rc p d k eku gks ld rk gS&
(A*) 0 (B*) 1 (C*) 1/2 (D) 3/5 Sol. The two engines have the same probabil ity of fail ing if nks bat u leku izkf;d rk d s gS 5C3p3(1 � p)2 + 5C4p4(1�p) + p5 = 3C2p2(1�p) + p3 10p3(1�p)25 + p4(1 � p) + p5 = 3p2(1 � p) + p3 2p3
� 5p2 + 4p � 1 = 0 p = 0 or ;k p = ½ or ;k p = 1
7. Let [.] denotes the greatest integer function and f(x) =
3
sin[x]a , x 0
x2 , x 0
sinx xb , x 0
x
.
If f(x) is continuous at x = 0 , then b is equal to : [CD-MS] [301]
length of chord t hok d h y EckbZ = |x1 � x2 | 21 m
= 24a 4a 1 4 = 40
i.e. ;k a2 + a � 2 = 0 i.e. ;k (a + 2) (a � 1) = 0
i.e. ;k a = � 2, 1.
Paragraph for Question Nos. 9 to 10 (iz'u 9 ls 10 ds fy, vuqPNsn)
Let A,B,C be the three points on the ellipse 2 2
2 2
x y
a b = 1 having eccentric angles
respectively. [EL-EA] [301]
ekuk A,B,C nh?kZoÙk 2 2
2 2
x y
a b = 1 ij rhu fcUnq gSA ft ud s mRd sUnz d ks.k gSA
9. Area of triangle PQR formed by corresponding points on auxiliary circle is JPT-2 Adv. 2015
(A*) ab
(area ABC) (B) ba
(area ABC) (C) 2ab
(area ABC) (D) (area ABC)
lgk;d oÙk ij l axr fcUnqvksa l s cuk f=kHkqt PQR d k {ks=kQ y gS&
(A*) ab
(ABC d k {ks=kQ y ) (B) ba
(ABC d k {ks=kQ y )
(C) 2ab
(ABC d k {ks=kQ y ) (D) (ABC d k {ks=kQ y )
Sol. Let ABC is formed by the points A(a cos, bsin), B(a cos, bsin) and C(a cos, bsin) on 2 2
2 2
x y
a b = 1
ekuk nh?kZoÙk 2 2
2 2
x y
a b = 1 ] A(a cos, bsin), B(a cos, bsin) and C(a cos, bsin) fcUnqvksa l s
,d f=kHkqt cuk;k t krk gSA
Let ekuk P(a cos, asin) Q(a cos, asin) ; R(a cos, asin) be point on auxiliary circle. Q(a cos, asin) ; R(a cos, asin) lgk;d oÙk ij gSA
Area of ABC d k {ks=kQ y = 12
acos bsin 1
acos bsin 1
acos bsin 1
= 12
ab
cos sin 1
cos sin 1
cos sin 1
area of PQR d k {ks=kQ y = 12
a2
cos sin 1
cos sin 1
cos sin 1
Area of ABC d k {ks=kQ y = ba
Area of PQR d k {ks=kQ y .
10. The eccentric angles of the vertices of triangle of maximum area inscribed in an ellipse differ by nh?kZoÙk d s vUrxZr vf/kd re {ks=kQ y d s f=kHkqt d s 'kh"kZ d s mRd sUnz d ks.kksa d k vUrj gS&
Paragraph for Question Nos. 11 to 12 (iz'u 11 ls 12 ds fy, vuqPNsn)
Let < an> and < bn> be the arithmetic sequences each with common difference 2 such that a1 < b1
and let cn =
n
1kka , dn =
n
1kkb . Suppose that the points An(an, cn), Bn(bn, dn) are all lying on the
parabola C: y = px2 + qx + r where p,q,r are constants. [SS-MS] [303] ekuk < an> vkSj < bn> lekUrj vuqØ e gS ft uesa izR;sd d k lkoZvUrj 2 bl izd kj gS fd
a1 < b1 vkSj ekuk cn =
n
1kka , dn =
n
1kkb . ekukfd fcUnq An(an, cn), Bn(bn, dn) ijoy ;
C: y = px2 + qx + r ij fLFkr gS t gk¡ p,q,r vpj gSA
11. The value of p equals p d k eku cjkcj gS&
(A*) 41
(B) 31
(C) 21
(D) 2
12. If r = 0 then the value of a1 and b1 are
(A) 21
and 1 (B) 1 and 23
(C*) 0 and 2 (D) 21
and 2
;fn r = 0 gks] rks a1 vkSj b1 d s eku gS&
(A) 21
vkSj 1 (B) 1 vkSj 23
(C*) 0 vkSj 2 (D) 21
vkSj 2
Sol. 17. Given fn;k x;k gS cn =a1 + a2 + a3 + �.. + an where a1, a2, ��., an are in A.P. with d = 2 t gk¡ a1, a2, ��., an l ekUrj Js<h esa gS ft ld k lkoZvUrj d = 2 gSA and dn = b1 + b2 + b3 + ���+ bn are in A.P. with d = 2 vkSj dn = b1 + b2 + b3 + ���+ bn lekUrj Js<h esa gSA ft ld k lkoZvUrj d = 2 gSA also (an, cn) lies on y = px2 + qx + r rFkk (an, cn) oØ y = px2 + qx + r ij fLFkr gSA Nowvc cn = pan + qan + r �..(1) cn�1 = pa2
n�1 + qa n�1 + r �.(2) From (1) and (2), we get (1) vkSj (2) ls gesa izkIr gksrk gSA cn � cn�1 = p(a2
n � a2
n�1) + q(an �an�1) an = p(an + an�1) (an �an�1) + q(an �an�1) an = (an � an�1) [p(an + an�1)+q] �.(3) (an � an�1 = d) on putting n = 2 and 3 in equation (3), we get n = 2 vkSj 3 esa j[kus ij (3), we get a2 = d[p(a2 + a1) + q] �.(4) a3 = d[p(a3 + a2) + q] �.(5) Now vc (5) � (4) , we get 3 2 3 1
Paragraph for Question Nos. 13 to 14 (iz'u 13 ls 14 ds fy, vuqPNsn)
If f : R R be a differentiable function such that (f(x))7 = x � f(x) then [AR-MS] [307] ;fn f : R R vod y uh; Q y u bl izd kj (f(x))7 = x � f(x) rc
13. The value of 2
0
1� dx)x(f is
2
0
1� dx)x(f d k eku gS&
(A) 3 2 (B) 2 (C*) 3 (D) 1 14. The area bounded by curve y = f(x) between the ordinates x = 0, x = 3 and x-axis is
(A*) 7f( 3 )
8 3 � f 3 � 4f( 3)8
sq. units (B)
73f�38
8)3(f
sq. units
(C)
893
�3f3 sq. units (D) none of these
oØ y = f(x) d ksfV x = 0, x = 3 vkSj x-v{k ls ifjc) {ks=k d k {ks=kQ y gS&
(A*) 7f( 3 )
8 3 � f 3 � 4f( 3)8
oxZ bd kbZ (B)
73f�38
8)3(f
oxZ bd kbZ
(C)
893
�3f3 oxZ bd kbZ (D) buesa l s d ksbZ ugha
Sol. f(x)[f(x)6 + 1] = x f(0)[(f(0))6 + 1] = 0 f(0) = 0 and vkSj 7 (f(x))6.f(x) = 1 � f1(x) f(x)[7(f(x))6 + 1] = 1 f(x) > 0 x R Hence f(x) is increasing function x R vr% f(x) , x R ,d o/kZeku Q y u gSA so there exists an inverse of f(x) such that f�1(0) = 0 blfy , f(x) d k izfry kse fo|eku gS t cfd f�1(0) = 0 x7 + x = f�1(x)
Now, we know that vc gesa izkIr gksrk gS dx)x(fdx)x(f)a(f
0
1�a
0 = af(a)
Hence vr% )3(f3dx)x(fdx)x(f)3(f
0
1�3
0
f ( 3 )3
7
0 0
f(x)dx 3f( 3) � (x x)dx
)3(f4�3f�38
8)3(f
dx)x(f7
3
0
INTEGER TYPE - 6 15. A game is played with special fair cubic die which has one red side, two blue sides, and three
green sides. The result is the colour of the top side after the dice is been rolled. If the die is rolled repeatedly. The probability that second blue result occurs on or before the tenth roll can be
expressed in the form of p q
r
3 � 2
3 where p,q,r are positive integers then find the value of p + r � q.
,d [ksy ] ,d fof'k"V fu"i{kikrh ?kuh; ikls ls [ksy k t krk gS ft ld h ,d Hkqt k y ky ] nks Hkqt k,a uhy h rFkk rhu Hkqt k,a gjh gSA ikls d ks Q sd us d s ckn Å ijh Hkqt k ij jax ifj.kke gSA ;fn ikls d ks y xkrkj Q sad k t krk gS] rks 10 ckj esa ;k mll s igy s iklk Q sad us ij nwljh ckj
uhy k ifj.kke vkus d h izkf;d rk d ks p q
r
3 � 2
3 d s : i esa O;Dr fd ;k t krk gSA t gk¡ p,q,r /kukRed
iw.kk±d gS] rc p + r � q d k eku gS& [PR-BT] [303] [JPT-2_Adv. 2015] Ans. 07 Sol. Now 2nd blue result occur on or before the 10th roll is equivalent to the occurrence of blue face
atleast twice is 10 rolls 10oha ckj ;k mll s igy s iklk Q sad us ij nwljh ckj uhy k ifj.kke vkus d h izkf;d rk] 10 ckj
esa d e l s d e nks ckj uhy h lrg vkus d s rqY; gSA
n = 10, p =13
q =23
P(r 2) = 1 � (P(0) + p(1))
= 1 � 10 9
2 1 210.
3 3 3
1 � 9 11 9 11
9 9 9
2 .4 2 3 � 21
3 3 3
p = 9 , q = 11, r = 9 p + r � q = 9 + 9 � 11 = 7
16. Let y be an element of the set A = {1, 2, 3, 5, 6, 10, 15, 30} and x1 x2 x3 = y, then the number of
positive integral solution of x1x2x3 = y is m. Find m. [PC-MT] [304] [JPT-2_Adv. 2015]
ekuk l eqPp; A = {1, 2, 3, 5, 6, 10, 15, 30} d k vo;o y gS vkSj x1 x2 x3 = y, rc x1x2x3 = y d s gy ksa
d s /kukRed iw.kk±d gy ksa d h la[;k m gSA m Kkr d hft ,A Ans. 64
Sol. The number of solutions of the given equation is the same as the number of solution of the equation x1x2x3 x4 = 30 = 2 × 3 × 5 (here x4 is dummy variable)
fn, x, lehd j.k d s gy ksa d h la[;k] lehd j.k x1x2x3 x4 = 30 = 2 × 3 × 5 d s gy ksa d h la[;k d s
cjkcj gS (;gk¡ x4 xqIr pj gS) Hence number of solutions is 43 = 64 vr% gy ksa d h la[;k 43 = 64 17. If area of triangle formed by the points (2, ),(+ , 2+ ) and (2, 2) is 8 square unit then
area of triangle whose vertex are ( + , � ), (3 � , + 3) and (3 � , 3 � ) is ;fn (2, ),(+ , 2+ ) vkSj (2, 2) ls cus f=kHkqt d k {ks=kQ y 8 oxZ bd kbZ gS rFkk
( + , � ), (3 � , + 3) vkSj (3 � , 3 � ) ls cus f=kHkqt d k {ks=kQ y gS& [JA_CT-2_21-09-2014] [JPT-2_Adv. 2015] [SL-AR] [301]
Ans.32
Sol.
P(2 ,2 )b aR(2 , )a b
C(3 � ,3 � )b b aaQ( + ,2 + )b b aa
D(3 � , +3 )b a b a
Area of ABC = 4(PQR) ABC d k {ks=kQ y = 4(PQR) = 4 × 8 = 32 square unit oxZ bd kbZ 18. If A1, A2, A3, A4 be the area of a triangular faces of a tetrahedron and h1, h2, h3, h4 be the
corresponding altitude of the tetrahedron. If volume of tetrahedron is 5 cubic units then find the
minimum value of 1 2 3 4 1 2 3 4(A A A A )(h h h h )
24
[SS-IG] [304] [JPT-2_Adv.
2015] ;fn A1, A2, A3, A4 ls cus prq"Q y d d s f=kHkqt kd kj lrgksa d k {ks=kQ y gS rFkk h1, h2, h3, h4
prq"Q y d d s l axr 'kh"kZ y Ec gSA ;fn prq"Q y d d k vk;ru 5 ?ku bd kbZ gS rc
1 2 3 4 1 2 3 4(A A A A )(h h h h )
24
d k U;wure eku gS&
Ans. 10
Sol. V = 13
h1A1 h1 =1
3VA
h2 = 2
3VA
h3 = 3
3VA
h4 = 4
3VA
(A1 + A2 + A3 + A4) 1 2 3 4
3V 3V 3V 3VA A A A
3V(A1 + A2 + A3 + A4) 1 2 3 4
1 1 1 1A A A A
1 2 3 4
1 2 3 4
A A A A 41 1 1 14A A A A
(A1 + A2 + A3 + A4) 1 2 3 4
1 1 1 1A A A A
16
Min value U;wure eku : 3V(16) = 48 V = 240
min value of U;wure eku 1 2 3 4 1 2 3 4(A A A A )(h h h h )
19. If roots of the cubic equation (z � ab)3 = a3, a 0 represents the vertex of a triangle, then the length of one of the sides of the triangle will be , when a = 3 [JPT-2_Adv. 2015]
;fn ?kuh; l ehd j.k (z � ab)3 = a3, a 0 d s ewy f=kHkqt d s 'kh"kZ gS] rc f=kHkqt d h Hkqt kvksa d s
,d d h y EckbZ gksxh] t c a = 3 [CN-CR] [302] Ans. 03 Sol. (z � ab)3 = a3
z � ab = a(1)1/3, z � ab = a, a, a2 z1 = ab + a, z2 = ab + a, z3 = ab + a2 |z1 � z2| = |a � a| = |a||1�| = 3 | a |
= 3 . 3 = 3
20. If the value of the definite integral = 3
6 3 21
(2x � 1)dx
x 2x 9x 1
can be expressed in the form of
�1A Ccot
B Dwhere A,B,C,D are rationals in their lowest form then the value of 2(A + B + C + D) is
equal to [DI-II] [301]
;fn fuf'pr lekd y = 3
6 3 21
(2x � 1)dx
x 2x 9x 1
d s eku d ks �1A Ccot
B Dd s : i esa O;Dr d jrs gS t gk¡
A,B,C,D mud s ljy re : i esa ifjes; la[;k,a gS] rc 2(A + B + C + D) d k eku cjkcj gS& Ans. 18
Sol. Here stretching force is same but stress will be different for wires. ;gk¡ ruu cy leku gS] ijUrq nksuksa rkjksa esa izfrcy fHkUu&fHkUu gSA For upper wire Åijh rkj ds fy,
22*. A particle of mass 'm' is suspended with the help of an ideal string from the ceiling of a car as
shown in figure. Initially the whole system is at rest. At time t = 0 the car starts accelerating towards right with uniform acceleration 'a' such that a << g. Then choose the correct statement(s).
[CM-VT](104) ,d vkn'kZ jLlh dh lgk;rk ls 'm' nzO;eku dk d.k fp=kkuqlkj dkj dh Nr ls yVdk gqvk gSA izkjEHk esa
lEiw.kZ fudk; fLFkj gSA t = 0 ij dkj nka;h vksj le:i Roj.k a ls Rofjr gksrh gS ,oa a << g gSA rks lgh dFku@dFkuksa dk p;u dhft,A
m
(A*) The particle will execute simple harmonic motion with respect to car. d.k dkj ds lkis{k ljy vkorZ xfr djsxk (B*) The particle will be at equilibrium position w.r.t. car for the 2nd time at instant
t = 2 2
32 a g
dkj ds lkis{k d.k t = 2 2
32 a g
le; i'pkr~ okil ek/; fLFkfr ij nwljh ckj igqprk gSA
(C) The maximum speed of the particle with respect to ground during subsequent motion will be
2 2 1/ 2(g a ) tan�1 ag
v/kksfyf[kr xfr ds nkSjku tehu ds lkis{k d.k dh vf/kdre pky 2 2 1/ 2(g a ) tan�1 ag
gSA
(D*) The speed of the particle with respect to ground at the instant the string becomes vertical
again for the first time is aT where T = 22 2a g
tc jLlh nqckjk izFke ckj Å /okZ/kj fLFkfr esa vkrh gS rc d.k dh tehu ds lkis{k pky aT gSa ;gk¡ T = 2
2 2a g
gSA
Sol. The particle will execute S.H.M with angular amplitude tan�1 ag
with respect to car and of time
period T = 22 2a g
.
d.k dkj ds lkis{k dks.kh; vk;ke tan�1 ag
rFkk vkorZdky T = 22 2a g
ds lkFk ljy vkorZ xfr djrk gSA
It will be at eqb. position for second time at time t = 3
Also, about the axis passing through A ; A ls xqtjus okyh v{k ds ifjr%
fluid tension
a b
a
gh bdh(h a)
= Vgx
x = m06.1V
dh)ah(hba
a
24. A solid sphere of mass m is released on the plank of mass M which lies on an inclined plane of
inclination as shown in figure. There is sufficient frictional force between sphere and plank and the minimum value of co-efficient of friction between plank and surfaces is to keep the plank at rest. Then [RB-CD](104)
(A*) Frictional force between sphere and plank is 27
mg sin , when plank is at rest
xksys rFkk r[rs ds e/; ?k"kZ.k cy 27
mg sin gS, tc r[rk fLFkj gSA
(B*) The value of is 7M 2m7(M m)
tan
dk eku 7M 2m7(M m)
tan gSA
(C) If there is no friction between the plank and inclined plane, then acceleration of plank is less than g sin . ;fn r[rs rFkk urry ds e/; dksbZ ?k"kZ.k cy vkjksfir ugha gks rks r[rs dk Roj.k g sin ls de gSA
(D*) If there is no friction between plank and inclined plane, then friction force on the sphere is zero.
;fn r[rs rFkk urry ds e/; dksbZ ?k"kZ.k cy vkjksfir ugha gks rks xksys ij vkjksfir ?k"kZ.k cy 'kwU; gSA Sol. If plank is at rest acceleration of cylinder ;fn r[rk fLFkj gks rks xksys dk Roj.k
25. Consider an infinite mesh as shown in figure Each side of the mesh has resistance R. Consider
hexagonal part ABCDEF of infinite mesh and equivalent resistance between any two points of hexagon is measured. Choose the correct option(s) : [CE-EQ](104)
(B*) equivalent resistance between A and C is R A rFkk C ds e/; rqY; izfrjks/k R gSA
(C*) equivalent resistance between A and D is 7R6
A rFkk D ds e/; rqY; izfrjks/k 7R6
gSA
(D*) equivalent resistance between A and E is R A rFkk E ds e/; rqY; izfrjks/k R gSA Sol. If we connect positive terminal of a battery to A and negative to infinity then the current through
branches will be as shown. If we connect negative to B and positive to infinity current will be in
opposite direction by supper -position of two situation net current in AB branch will be 23
;fn cSVjh ds /kukRed fljs dks A ls rFkk _ .kkRed fljs dks vuUr ls tksM+ ns rks bl ifjiFk dh 'kk[kkvksa esa /kkjk fp=kkuqlkj izokfgr gksxhA ;fn cSVjh ds _ .kkRed fljs dks B ls rFkk /kukRed fljs dks vuUr ls tksM+ ns rks /kkjk
foifjr fn'kk esa izkokfgr gksxh] rFkk bu nksuksa fLFkfr;ksa esa izokfgr /kkjk ds v/;kjksi.k }kjk 'kk[kk AB esa dqy /kkjk 23
izokfgr gksxhA A rFkk B ds e/; foHkokUrj 2
R3
gksxkA blfy, A rFkk B ds e/; rqY; izfrjks/k 2R3
gksxk] vkSj blh
izdkj vU; fljks ds fy, Hkh Kkr dj ldrs gSA
A
/3
/6 /6
/3 /3 /6 /6
26. Consider an imaginary cube of side '' as shown in figure : [ES-GT](103) fp=k esa iznf'kZr '' Hkqtk ds ,d dkYifud ?ku dh dYiuk dhft, :
A
E
C
G
B
F
D
H
Q
P
An infinite large sheet of surface charge density passes through PQGF. P and Q are mid points
of AB and DC respectively. Out of six faces of cube the electric flux linked : i"Bh; vkos'k ?kuRo dh ,d vUur vkdkj dh ifV~Vdk PQGF ls xqtjrh gSA P rFkk Q Øe'k% AB rFkk DC
ds e/; fcUnq gSA blds N% Qydksa ls lEcfU/kr ¶yDl esa ls : (A*) is zero with three faces rhu Qydksa ds fy, 'kwU; gSA (B*) with face ADHE and face BCGF are equal Qyd ADHE rFkk Qyd BCGF ds fy, leku gSA (C*) is minimum (non zero) with FGHE FGHE ds fy, U;wure (v'kwU;) gSA (D) is zero with two faces nks Qydksa ds fy, 'kwU; gSA Sol. Consider the projected area of any face to calculate the flux ¶yDl Kkr djus ds fy, fdlh Qyd ds {kS=kQy dh dYiuk djrs gSA Electric flux with ABCD, ABEF, DCGH is zero ABCD, ABEF, DCGH ls lEcfU/kr fo|qr ¶yDl 'kwU; gSA
27. Figure shows top view of a horizontal surface. Two blocks each of mass m are placed on the surface and connected with a string. The friction coefficient is for each block. A horizontal force F is applied on one of the block as shown in the figure. F is maximum so that there is no sliding at any contact. [FR-SG](104)
,d {kSfrt lrg dk Å ijh n'; fp=k esa iznf'kZr gSA leku nzO;eku m ds nks CykWd bl lrg ij fLFkr gS rFkk vkil esa ,d jLlh ls cU/ksa gq, gSA izR;sd CykWd ds fy, ?k"kZ.k xq.kkad gSA fp=kkuqlkj ,d CykWd ij ,d {kSfrt cy F vkjksfir fd;k tkrk gSA F vf/kdre gS ftlds dkj.k fdlh Hkh lEidZ ij fQlyu ugha gSA
F
Top view
m
(A*) If ;fn = 30° F = 3
mg2 & T < mg
(B*) If ;fn = 45° F = mg2 & T = mg
(C) If ;fn = 60° F = 2mg & T < mg
(D*) If ;fn = 60° F = mg3 & T = mg Sol.
f T
F
T cos = f sin but T mg T cos = mg sin F = T sin + f cos
= mg
cossin
sin + mg cos
= mg
cos)cos(
For Fmax = < 45° Fmax = mg sec Fmax ds fy,] = < 45° Fmax = mg sec But tension in the string can not be more than mg jLlh esa ruko mg ls T;knk ugha gks ldrk
T = mg sin cos
< mg should be less than 45°
T = mg sin cos
< mg ] 45° de gksuk pkfg,A
Now if (vc ;fn) > 45° T = mg = 90° � F = mg sin + mg cos = 2mg sin If (;fn) < 45° then (rks) Fmax = mg sec
28. Consider a hemisphere of radius R with centre of curvature at origin O, as shown. Refractive index
of material of the hemisphere varies as = 2R
2R x. Where x is x-coordinate of material point. A
ray travelling in air in xy-plane is grazingly incident at O, as shown R f=kT;k ds ,d v)Zxksys dh dYiuk dhft,] ftldk oØrk dsUnz ewyfcUnq O ij fp=kkuqlkj gSA v)Zxksys ds
inkFkZ dk viorZukad = 2R
2R x ds vuqlkj ifjofrZr gksrk gSA tgk¡ x inkFkZ ds fcUnq dk x-funsZ'kkad gSA ,d
fdj.k ok;q esa xfr djrh gqbZ] fp=kkuqlkj xy-ry esa fcUnq O ij i"BLi'khZ vkifrr gksrh gSA [GO-RP](105) y
O x
(A*) Trajectory followed by ray as it travels inside the hemisphere is circular v)Z xksys eas xfr ds nkSjku fdj.k dk iFk oÙkkdkj gSA (B) y-coordinate of the point of hemisphere where the ray comes out of the hemisphere lies
between 0.5R and 0.75R fdj.k v)Zxksys ds ftl fcUnq ls ckgj fudyrh gS] v)Zxksys ds ml fcUnq dk y-funsZ'kkad 0.5R ls 0.75R ds
e/; fLFkr gSA (C*) Deviation suffered by the ray just before it comes out of the hemispherical surface lies
between 0° and 30° xksyh; lrg ls Bhd ckgj fudyrs le; fdj.k }kjk izkIr fopyu 0° ls 30° ds e/; fLFkr gSA (D) Deviation suffered by the ray just before it comes out of the hemispherical surface lies between
30° and 45° xksyh; lrg ls Bhd ckgj fudyrs le; fdj.k }kjk izkIr fopyu 30° ls 45° ds e/; fLFkr gSA Sol.
O
x 90°
A
The figure shows a strip at a distance x of thickness dx. As of material increases with x, ray will
deviate continuously as shown. By snell's law, between O and A, fp=k esa x nwjh ij dx eksVkbZ dh ifêdk iznf'kZr gSA pqafd nwjh x ds lkFk inkFkZ dk viorZukad c<+rk gS] bl
dkj.k fdj.k fp=kkuqlkj yxkrkj fopfyr gksrh gSA O ls A rd Lusy ds fu;e }kjk
y2 + (x-2R)2 = (2R)2 which is equation of circle of radius 2R centered at (2R, 0) tksfd ,d 2R f=kT;k ds oÙk dh lehdj.k gS ftldk dsUnz (2R, 0) ij gSA also, equation of hemispherical surface is blh izdkj v)Zxksys dh lrg dh lehdj.k fuEu gS x2 + y2 = (R)2 y2 = R2 � x2 Putting this value of y2 in equation of trajectory we can find coordinates of point where the ray
comes out of hemisphere, as bl y2 ds eku dks iFk lehdj.k esa j[kus ij mu fcUnqvksa ds funsZ'kkad dj ldrs gS tgk¡ ij fdj.k v)Zxksys ls
ckgj vkrh gS R2 + x2 + (x � 2R)2 = 4R2
x = R4
y = 2
2 2 2 R 15RR x R 0.97R
4 4
Now, to find deviation () consider the diagram here vc, fopyu () Kkr djus ds fy, fuEu fp=k dk iz;ksx djrs gSA
y
y 2R
Clearly Li"V :i ls, sin =
y 15 16 1or
2R 8 8 2
0 < < 30°
SECTION � 2 : (Paragraph Type)
[k.M � 2 : (vuqPNsn çdkj)
This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relate to three paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).
bl [k.M esa fl)karksa] ç;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSaA rhuksa vuqPNsnksa ls lacaf/kr N%
ç'u gSa] ftuesa ls gj vuqPNsn ij nks ç'u gSaA fdlh Hkh vuqPNsn esa gj ç'u ds pkj fodYi (A), (B), (C) vkSj
(D) gSa] ftuesa ls dsoy ,d gh lgh gSA
Paragraph for Question Nos. 29 to 30 iz'u 29 ls 30 ds fy, vuqPNsn
Comp.
The rectangular box shown in the figure has a partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a monatomic ideal gas at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat loss through lead wire of heater is negligible. The gas in left chamber expands, pushing the partition until final
pressure in both chambers becomes 24332
P0. (Take R = 8.3 J/mol K) [TH-FL](104)
fp=k esa iznf'kZr ?kukdkj fMCcs esa fLFkr fiLVu fcuk ?k"kZ.k ds fMCcs dh yEckbZ ds vuqfn'k fQly ldrk gSA izkjEHk esa fMCcs ds nksuksa Hkkxksa esa ,d eksy ,dijek.kqd vkn'kZ xSl] nkc P0, vk;ru V0 rFkk rkieku T0 ij Hkjh gqbZ gSA cka;s okys Hkkx esa fLFkr xSl dks fo|qr ghVj }kjk /khjs&/khjs xeZ fd;k tkrk gSA fMCcs dh nhokjs rFkk fiLVu Å "eh; dqpkyd gS rFkk ghVj ds rkj }kjk Å "eh; gkfu ux.; gSA cka;s okys Hkkx esa fLFkr xSl QSyrh gS
29. Final temperature of gas in left chamber is : [TH-FL](104) cka;s Hkkx esa xSl dk vfUre rkieku gksxk & (A) 2.25 T0 (B) 4.5 T0 (C) 8.75 T0 (D*) 12.93 T0
Sol. 0
right
T
T
= 1
0
0
P 32
243P
0
right
T
T =
223
Tright = 2.25 T0
For the left chamber, the process of heating is slow. Initial parameters are P0, V0, T0. nka;s Hkkx ds fy,, izØe vuqlkj /khsjs&/khjs xeZ djrs gSA izkjfEHkd izkapy P0, V0, T0 gSA
Final parameters are vfUre izkapy 0243
P32
, V1, T1
0 0
0
P V
T = 0243P
32
1
1
VT
T1 = 24332
1
0
VV
T0 ���.(1)
Adiabatic compression occurs in right chamber. Initial parameters are P0, V0, T0. nka;s Hkkx esa :)ks"e lEihM+u gksxkA blds izkjfEHkd izkapy P0, V0, T0 gSA
by equation lehdj.k (1) & o (2) ls T1 = 12.93 T0 30. The work done by the gas in the left chamber is [TH-FL](104) cka;s Hkkx esa xSl }kjk fd;k x;k dk;Z gksxk (A) 5.58 T0 J (B) 10.58 T0 J (C) 25.58 T0 J (D*) 15.58 T0 J Sol. Work done by gas on right chamber. nka;s Hkkx esa fLFkr xSl }kjk fd;k x;k dk;ZA
W = 2 2 0 0P V P V
1
= 32
9
14
P0V0
= 158
× 8.3 T0
= (�15.58 T0) J
Paragraph for Question Nos. 31 to 32
iz'u 31 ls 32 ds fy, vuqPNsn Comprehension # 1 In standard YDSE the intensity on screen varies with position according to following graph as we
move away from the central maxima : ] ekud YDSE iz;ksx esa dsfUnz; mfPp"B ls nwj tkus ij fLFkfr ds lkFk insZ ij izkIr rhozrk fuEu vkjs[k ds
vuqlkj ifjofrZr gksrh gSA [YE-YE](104)
0
2 4 6 8 10 y (in mm)
0
Wave length of light used is 400 nm and distance between the two slits is 1 mm. iz;qDr izdk'k dh rjaxnS/;Z 400 nm rFkk nksuksa fLyVksa ds e/; nwjh 1 mm. gSA
eksVkbZ okyh iryh fQYe j[kus ij y = 2 mm ij rhozrk D;k gksxh ? [YE-YE](104)
(A*) /2 (B) (C) /4 (D) zero
Sol. Shift foLFkkiu = t( 1)
= �6
7
2.2 10 0.5
4 10
= 114
So, intensity will become 0
2
vr% rhozrk 0
2
gks tk;sxhA
Paragraph for Question Nos. 33 to 34 iz'u 33 ls 34 ds fy, vuqPNsn
Passage The common prism telescope is shown in figure is generally made up of three parts, viz., a prism and two
telescopes. One of these telescopes is used to render the rays which fall on the prism parallel to each other, and is therefore called the collimator, the other telescope is made movable so that it can be turned into the proper direction for observing any desired color, and is, therefore, called the view-telescope. A circular scale is also provided to major angular position of telescope.
On the circular scale 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (0.5°). [GO-OI](104)
First telescope is focused to collect rays directly from collimator and position of telescope is measured. This is position (1). Now a prism is placed as shown in the figure and telescope is focused for red, yellow and violet. The respective positions of telescope are (2), (3) and (4).
33. What is deviation for violet light : [GO-OI](104) cSaxuh jax ds fy, fopyu D;k gksxk % (A) 12°22 (B) 13°19 (C) 33°5 (D*) 13°21 Sol. 29 division of main scale coincides with 30 divisions of vernier scale Hence one division of vernier
DV = 33°41� 20°20= 13°21 34. If we find angular derivation for different colours then percentage error in derivation will be
maximum for [GO-OI](104) (A*) Red (B) Yellow (C) Violet (D) Same for all ;fn ge vyx jaxksa ds fy, dks.kh; fopyu Kkr djs rks izfr'kr fopyu =kqfV fdl jax ds fy, T;knk gksxhA (A) yky (B) ihyk (C) cSaxuh (D) lHkh ds fy, leku gSA
Sol. Percentage error izfr'kr =kqfV =
× 100
is minimum for red is same for all, so percentage error will be maximum for Red. yky jax ds fy, U;wure gS lHkh ds fy, leku gS, vr% yky jax ds fy, izfr'kr =kqfV vf/kdre gksxhA
35. A particle starts from rest and moves on a circular path with an angular acceleration which is proportional to square root of instantaneous angular velocity. Find the mean angular velocity (in rad/sec) averaged over the time it takes to acquire an angular velocity of 12 rad sec�1
,d d.k fLFkj voLFkk ls oÙkkdkj iFk ij dks.kh; Roj.k ds lkFk xfr izkjEHk djrk gSA bldk dks.kh; Roj.k
blds rkR{kf.kd dks.kh; osx ds oxZewy ds lekuqikrh gSA 'kwU; ls 12 rad sec�1 dks.kh; osx izkIr djus esa yxs le;kUrjky ds nkSjku vkSlr :i ls bldk ek/; dks.kh; osx (rad/sec esa) Kkr djksA [CM-KN](104)
36. A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 37C it
weighs 30 g. When the temperature of the liquid is raised to 42C, the metal piece weighs 30.5 g. The specific gravity of the liquid at 42C is 1.20. Coefficient of linear expansion of the metal is nearly 69 × 10
�N, find N. [FL-BY](104)
/kkrq ds ,d VqdM+s dk gok esa Hkkj 46 g gSA tc bldks 37C rkieku ij 1.24 fof'k"V xq:Ro ds nzo esa Mqcks;k
40. A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. The difference between the
pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure is 10x N/m2. Surface tension of water = 0.075 N/m contact angle between glass and water is zero degree. Find the value of x. (g = 9.8 m/s2) [SF-EX](103)
0.50 feeh f=kT;k dh ds'kuyh dks ty ds ik=k esa m/okZ/kj Mqck;k x;k gSA uyh esa p<s gq, ty dh lrg ls 5.0 lseh uhps nkc ,oa ok;qe.Myh; nkc dk vUrj 10x N/m2 gks rks x dk eku Kkr dhft;sA ty dk i"B rukao =
0.075 N/m rFkk dkWp o ikuh ds e/; Li'kZ dks.k 'kwU; fMxzh gSA (g = 9.8 m/s2) Ans. 19 Sol.
Course : JR AIOT (JEE ADVANCE) - 2 Test Date : 03-05-2015
Test Type : JEE ADVANCED -2 (ELPD)
Page # 1
PAPER-2 MCQ (6) 41. A 5g sample containing Fe3O4 (FeO + Fe2O3) and an inert impurity is treated with excess of KI solution in
the presence of dilute H2SO4. The entire iron converted to Ferrous ion along with liberation of Iodine. The resulting solution is diluted to 100 ml. 20 ml of the diluted solution requires 10 ml of 0.5M Na2S2O3 solution to reduce the iodine present. Amongs the following select correct statements.
(A*) % of Fe2O3 in sample is 40% (B) % of FeO in sample is 28% (C*) % of inert impurity in sample is 42% (D) % of inert impurity in sample is 32% Fe3O4 (FeO + Fe2O3) rFkk ,d vfØ; v'kqf);qDr] ,d 5g izkn'kZ dk ruq H2SO4 dh mifLFkfr esa KI foy;u ds vkf/kD;
0.5M Na2S2O3 dk 10ml vko';d gksrk gSA fuEu esa ls lgh dFkuksa dk p;u dhft,A (A*) izkn'kZ esa Fe2O3 dk 40% gSA (B) izkn'kZ esa FeO dk 28% gSA (C*) izkn'kZ esa vfØ; v'kqf) 42% gSA (D) izkn'kZ esa vfØ; v'kqf) 32% gSA Sol. 40 Let the moles of Fe3O4 is x. So mole of Fe2O3 (in Fe3O4) = x eq of Fe2O3 (in Fe3O4) = eq of K = eq of 2 = eq of Na2S2O3. or 2x = 0.025 or x = 0.0125 mol of Fe2O3 So mass of Fe2O3 = 0.0125 × 160 = 2g
% of Fe2O3 = 25
× 100 = 40%.
Sol. 40 ekuk fd Fe3O4 ds eksy x gSA vr% Fe2O3 ds eksy (Fe3O4 esa) = x
42. -bond results due to overlap of : (CBO-VBT)_203 (A*) dxy and py along x�axis (B) dx2�y2 and py along x�axis (C*) dxy and px along y�axis (D) dx2�y2 and py along y�axis
-cU/k] fuEu ds vfrO;kiu dk ifj.kke gksrk gS % (A*) x-v{k dh vksj dxy rFkk py ds vfrO;kiu ds dkj.k (B) x-v{k dh vksj dx2�y2 rFkk py ds vfrO;kiu ds dkj.k (C*) y-v{k dh vksj dxy rFkk px ds vfrO;kiu ds dkj.k (D) y-v{k dh vksj dx2�y2 rFkk py ds vfrO;kiu ds dkj.k Sol. -bond are formed by parallel overlaping of same orbitals. So, correct options are : (A) dxy and py along x�axis (C) dxy and px along y�axis
gy leku d{kdksa ds lekukUrj ¼ikf'oZd½ vfrO;kiu ds dkj.k - ca/k dk fuekZ.k gksrk gSA vr% lgh fodYi gSa % (A) x-v{k dh vksj dxy rFkk py ds vfrO;kiu ds dkj.k (C) y-v{k dh vksj dxy rFkk px ds vfrO;kiu ds dkj.k 43. The vapour pressure of two miscible liquids A and B are 300 and 500 mm of Hg respectively. In a flask,
2 moles of A are mixed with 6 moles of B. Further to the mixture, 32 g of an ionic non-volatile solute MCl (partially ionised, mol. mass = 70 u) were also added. Thus, the final vapour pressure of solution was found to be 420 mm of Hg. Then, identify the correct statement(s) : (Assume the liquid mixture of A and B to behave ideally). (SCP-RLVP)_206
(A*) The numerical value of relative lowering in vapour pressure upon addition of solute MCl is 1/15.
Page # 2
(B*) The solute MCl is 25% ionised in the above question. (C) The solute MCl is 23.33% ionised in the above question. (D*) Upon addition of excess Pb(NO3)2, the number of moles of PbCl2 precipitated is 2/35.
nks feJ.kh; nzoksa A rFkk B ds ok"i nkc Øe'k% 300 rFkk 500 mm Hg gSaA ,d ¶ykLd esa] 2 eksy A dks B ds 6 eksy ds lkFk fefJr fd;k tkrk gSA blds ckn bl feJ.k esa] ,d vok"i'khy vk;fud foys; MCl ¼vkaf'kd :i ls vk;fur] vkf.od nzO;eku = 70 u½ dk 32 g Hkh feyk;k tkrk gSA bl çdkj foy;u dk vfUre ok"i nkc 420 mm Hg ik;k tkrk gSA rc lgh dFku@dFkuksa dh igpku dhft, % (A rFkk B ds nzo feJ.k dk O;ogkj vkn'kZ ekusaA)
44. Which of the following properties is/are common to both K2Cr2O7 & KMnO4 ? (DBC-IDM)_203 (A*) Both on heating form an oxosalt, a metal oxide & oxygen gas. (B*) Both have their colour because of charge transfer spectrum. (C) Both behave as primary titrants as well as self-indicators in redox titrations. (D*) Both form a slightly coloured precipitate with hydrogen sulphide in acidic medium. fuEu esa ls dkSulk@dkSuls xq.k/keZ K2Cr2O7 rFkk KMnO4 nksuks ds fy, leku gS@gSa \
(A*) nksuksa dks xeZ djus ij vkDlks yo.k] /kkrq vkWDlkbM o vkWDlhtu xSl curh gSA
(A*) F reacts with KMnO4/OH gives phthalic acid which upon heating gives phthalic anhydride.
(B*) E upon monochlorination (Cl2/h) gives total four products. (C*) F upon complete catalytic hydrogenation gives saturated compound which show geometrical
isomerism. (D*) Compound (A) gives the NaHCO3 test.
(A*) ;kSfxd F, KMnO4/OH ds lkFk fØ;k djds FkSfyd vEy cukrk gS ftls xeZ djus ij FkSfyd ,ugkbMªkbM curk gSA
(B*) ;kSfxd E ds ,dyDyksjksuhdj.k (Cl2/h) ij dqy pkj mRikn curs gSA
(C*) ;kSfxd F ds iw.kZ mRizsjdh; gkbMªkstuhdj.k ij larIr ;kSfxd curk gS tks T;kferh; leko;ork n'kkZrk gSA
(D*) ;kSfxd (A) NaHCO3 ijh{k.k nsrk gSA
Sol.
+
CH2�C
O
O CH2�C
O
3AlCl
C CH2
CH2 HO�C
O
O (A)
HCl
Hg�Zn
C
O (B)
H
5PCl
C
O (C)
Cl
AlCl3
O (D)
Zn�Hg + HCl
(E)
Pd�C/
Cl2/h
(d)
Cl H *
+
(d)
Cl H
*
(F)
KMnO4
Pd + H2 /
H
H + H
H
OH
COOH
COOH
�H2O
O
O
O
cis.
trans. (G)
Page # 5
48.
Product (X) Conc. H2SO4
80ºC Reaction-I
Conc. H2SO4
160ºC
Product (Y) Conc. H2SO4
160ºC Reaction-II
Which of the following option(s) is/are correct for above give reactions �
(SSS Sir) (AC-EAS) (MCQ) (204) (M)
(A*) In the reaction-I product is formed by most stable intermediate and product (X) is converted into (Y) by
conc. H2SO4 at 160ºC
(B) In the reaction-II product is -naphthalene sulphonic acid
(C*) Product 'X' is converted into product Y by conc.H2SO4/ and it is a thermodynamically controlled
process.
(D) In the reaction I & II electrophile attacks on naphthalene with same regioselectivity.
mRikn (X)
lkUnz H2SO4
80ºC vfHkfØ;k-I
lkUnz H2SO4
160ºC
mRikn (Y) lkUnz H2SO4
160ºC vfHkfØ;k-II
mijksDr nh xbZ vfHkfØ;kvksa ds fy, fuEu esa ls dkSuls fodYi lgh gS@gSa& (SSS Sir) (AC-EAS) (MCQ) (204) (M)
(A*) vfHkfØ;k-I esa mRikn lokZf/kd LFkk;h e/;orhZ ls curk gS rFkk mRikn (X) 160ºC ij lkUnz H2SO4 dh mifLFkfr esa
;kSfxd (Y) esa ifjofrZr gks tkrk gSA
(B) vfHkfØ;k-II esa mRikn -uS¶FkSyhu lY¶;wfjd vEy gksrk gSA
(C*) mRikn 'X' lkUnz H2SO4/ dh mifLFkfr esa mRikn Y esa ifjofrZr gks tkrk gS rFkk ;g vfHkfØ;k Å "ekxfrdh;
:i ls fu;af=kr izØe gSA
(D) vfHkfØ;k I o II esa bysDVªkWu Lusgh leku fjft;ksp;ukRedrk ds lkFk us¶Fksyhu ij vkØe.k djrk gSA
Sol.
SO3/80ºC
H SO3 �
�H
SO3H
SO3/160ºC
H
SO3 �
�H SO3H
Page # 6
Comp.(3 x 2Q.) (2) Paragraph for Question Nos. 49 to 50
iz'u 49 ls 50 ds fy, vuqPNsn Calomel electrode : It consists of mercury at the bottom over which a paste of mercury-mercurous
chloride is placed. A saturated solution of potassium chloride is then placed over the paste. A platinum wire sealed in a glass tube helps in making the electrical contact. The electrode is connected with the help of the side tube on the left through a salt bridge with the other electrode to make a complete cell.
Fig. Calomel electrode
The potential of the calomel electrode depends upon the concentration of the potassium chloride solution.
If potassium chloride solution is saturated, the electrode is known as saturated calomel electrode (SCE) and if the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) while for 0.1 N potassium chloride solution, the electrode is referred to as decinormal calomel electrode (DNCE). The electrode reaction when the electrode acts as cathode is :
12
Hg2Cl2 + e� Hg + Cl�
(Given : 2.303 RT / F = 0.06) Now answer the following questions. dSyksey bysDVªksM : blds vUrxZr edZjh rys ij gksrh gS ftlds Å ij edZjhµ ejD;wjl DyksjkbZM dk isLV j[kk tkrk gSA
The coordination number of iron in both the complexes (A) and (B) is six. (QUA-IIG)_204 Now answer the following questions.
Page # 8
nksuksa ladqyksa (A) o (B) esa vk;ju dh leUo; la[;k N% gSA vc fuEu iz'uksa ds mÙkj nhft,A 51. Which of the following options is incorrect ? (COR-ACFT)_203 (A) A + Fe2+ = Prussion blue colour (B) A + Fe3+ = Brown colour (C*) B + Fe2+ = Light green colour (D) B + Fe3+ = Prussion blue colour fuEu esa ls dkSulk fodYi lgh ugha gS \ (A) A + Fe2+ = izqf'k;u uhyk jax (B) A + Fe3+ = Hkwjk jax (C*) B + Fe2+ = gYdk gjk jax (D) B + Fe3+ = izqf'k;u uhyk jax Sol. B + Fe2+ = white colour B + Fe2+ = lQsn jax 52. Which statement is incorrect ? (COR-ACFT)_203 (A) Complex A absorbs radiation of greater frequency. (B) Complex B has smaller spin magnetic moment. (C) Both complexes have same hybridisation of central atom/ion. (D*) Complex B is thermodynamically more stable. fuEu esa ls dkSulk dFku xyr gS \ (A) ladqy A mPp vkorhZ dh fofdj.ksa vo'kksf"kr djrk gSA (B) ladqy B de pØ.k pqEcdh; vk?kw.kZ j[krk gSA (C) nksuksa ladqyksa esa dsUnzh; ijek.kq@vk;u dk ladj.k leku gksrk gSA
(D*) ladqy B m"ekxfrdh; :i ls vf/kd LFkk;h gksrk gSA Sol. (A) Complex A - K3[Fe(CN)6] absorbs radiation of greater frequency because it contains Fe3+, so greater
O. (B) Complex B - K4[Fe(CN)6] has zero spin magnetic moment. (C) Both complexes have same hybridisation (d2sp3) of central atom/ion. (D) Complex B is thermodynamically less stable (O.N. stability). (A) ladqy A - K3[Fe(CN)6] mPp vkorZ dh fofdj.ksa vo'kksf"kr djrk gSA D;ksafd ;g Fe3+ vk;u j[krk gSA blfy, O
mPp gksrk gSA (B) ladqy B - K4[Fe(CN)6] dk 'kwU; pØ.k pqEcdh; vk?kzw.kZ gksrk gSA (C) nksuksa ladqy ds dsUnzh; ijek.kq@vk;u dk ladj.k (d2sp3) gksrk gSA (D) ladqy B m"ekxfrdh; :i ls de LFkk;h gksrk gSA (vkWDlhdj.k vad LFkkf;Ro).
Paragraph for Question Nos. 53 to 54
iz'u 53 ls 54 ds fy, vuqPNsn Each of these reactions is a substitution of the leaving group (�OTs or �Cl) by solvent.
PhS
Cl 2
1
H O
r (r1 > r2)
Cl 2
2
H O
r
The mechanism by which they speed up the reactions is known as neighbouring group participation.
PhS
Cl Ph�S
H2O
Three membered ring intermediate
OH S
Ph
In this, ionization of the starting material is assisted by the lone pair of an electron-rich functional group.
(RSS Sir) (RM) (NGP) (Tough) 53. Which of the following are true statement -
(A)
OAc
OTs
reacts with acetic acid faster than
OAc
OTs
Page # 10
(B) O
Cl reacts with H2O slower than Cl
(C*)
SMe OSO2Ph Me reacts with ROH raster than
Me OSO2Ph Me
(D)
I HO reacts faster than
I HO with H2O.
fuEu esa ls dkSulk dFku lR; gS &
(A) ,flfVd vEy ds lkFk
OAc
OTs
dh fØ;k
OAc
OTs
dh rqyuk es vf/kd rsth ls gksrh gSA
(B) H2O ds lkFk O
Cl dh fØ;k Cl dh rqyuk es /kheh gksrh gSA
(C*) ROH ds lkFk
SMe OSO2Ph Me dh fØ;k
Me OSO2Ph Me dh rqyuk esa vf/kd rst gksrh gSA
(D)
I HO ,
I HO fd rqyuk esa H2O ds lkFk rsth ls vfHkfØ;k djrk gSA
Sol.
SMe OSO2Ph Me show neighbouring group participation.
SMe OSO2Ph Me iM+kSlh lewg dk ;ksxnku iznf'kZr djrk gSA
54.
OAc
OTs
AcOH Product, ¼mRikn½
Product is ¼mRikn gS½ �
(A)
OAc
OAc
(B*)
OAc
OAc
(C)
OAc
OTs
(D)
OAc
OAc
Sol. In
OAc
OAc
neighbouring group participation is not possible.
OAc
OAc
esa iM+kSlh lewg dk ;ksxnku lEHko ugh gSA
Integer (Double digit) (4)
55. [Au(CN)x]� is a very stable complex (though EAN = 82) under certain conditions and Kf of [Au(CN)x]
� is 4
1028. z 10�y M concentration of cyanide ion is required to maintain the equilibrium at which 99 mole % of the gold is in the form of cyanide complex. Then determine (y + z). Atomic number of Au = 79. z is a natural number & 1 z 9. (CEQ-HOEL, COR-BICC)_205
Sol. Equation of straight line (lh/kh js[kk dk lehdj.k esa) :
(y � y1) =2 1
2 1
y y
x x
(x � x1)
(P � 16) =4 1615 6
(V � 6)
3P + 4V = 72
Tmax =max(PV)
nR
For (PV)max ((PV)max ds fy,), 3P =722
and (rFkk) 4V =722
P = 12, V = 9
Tmax =12 9
2 (1/ 12)
= 648K.
KEmax = 32
nRTmax = 32
× 2 ×1
12× 648 = 162 L atm.
58. [H+] concentration in 0.01 M H2O2 solution (1aK = 3 × 10�12 and
2aK 0) is x M. Fill first two digits of 108 x
as answer. (IEQ-PPAB)_205
0.01 M H2O2 ( 1aK = 3 × 10�12 rFkk 2aK 0) foy;u esa [H+] lkanzrk x M gSA 108 x ds igys nks vadksa dks mÙkj ds :i
esa Hkfj,A
Ans. 20
Sol. [H+] = 12 2 14a 0 wK C K 3 10 10 10
= 2 × 10�7 M = x 108 x = 20
Page # 13
Integer (Double digit) (2)
59. How many cyclic structural isomer of C6H10 possible which after ozonolysis give diketones that show Intra molecular Aldol condensation. (Only between ketone).