JEE (MAIN+ADVANCED)

JEE ADV.2023

TARGET TEST PATTERN JEE (ADVANCED)

JEE (MAIN+ADVANCED)
EXAMINATION
ADVANCED MAJOR
TARGET YEAR 2024 TEST TYPE
TEST (AMT)
PAPER NO. ONE TEST CODE &
SEQUENCE
AMT 1
PAPER CODE 1 MAX. MARKS 180
CLASS XII TEST
3 Hrs
DURATION
COURSE NAME VIJETA TEST DATE 01ST DECEMBER 2023
COURSE CODE JP TEST DAY FRIDAY
TEST TIME Start: 09:30 AM
PHASE CODE(S) 01,02,03,05JP,JAZP,03JPAH
(PAPER-1) End : 12:30 PM
BATCH CODE(S) 01,02,03,05JP,JAZP,03JPAH TOTAL NO. OF
PAGES IN PAPER 44
BOOKLET

TEST PAPER DETAILS MARKING SCHEME

Full Partial Marks

Type No.
Subject Marks If Three If Two If One If No (–)ve Total Subject
Qs. No. Section No. of of
Sequence Per Options Options Option Option Marks Marks Total
Qs.* Qs.
Qs. Correct Correct Correct Chosen
1 to 03 1 MSQ 3 4 +3 +2 +1 0 –2 12
04 to 07 2 MCQ 4 3 - - - 0 –1 12
Maths 60
08 to 13 3 NVQ 6 4 - - - 0 0 24
14 to 17 4 MLQ 4 3 - - - 0 –1 12
18 to 20 1 MSQ 3 4 +3 +2 +1 0 –2 12
21 to 24 2 MCQ 4 3 - - - 0 –1 12
Physics 60
25 to 30 3 NVQ 6 4 - - - 0 0 24
31 to 34 4 MLQ 4 3 - - - 0 –1 12
35 to 37 1 MSQ 3 4 +3 +2 +1 0 –2 12
38 to 41 2 MCQ 4 3 - - - 0 –1 12
Chemistry 60
42 to 47 3 NVQ 6 4 - - - 0 0 24
48 to 51 4 MLQ 4 3 - - - 0 –1 12
Total 51 MAXIMUM MARKS 180
* Please turn overleaf to understand the meaning of coding for types of Questions.
A. GENERAL INSTRUCTIONS B. DARKENING THE BUBBLES ON THE ORS :
1. Use a BLACK BALL POINT to darken the
1. Darken the appropriate bubbles on the original
bubbles in the upper sheet.
by applying sufficient pressure. 2. Darken the bubble COMPLETELY.
2. The original is machine-gradable and will be 3. Darken the bubble ONLY if you are sure of the
collected by the invigilator at the end of the answer.
examination. 4. The correct way of darkening a bubble is as
3. Do not tamper with or mutilate the ORS. shown here :
5. There is NO way to erase or "un-darkened
4. Write your name, roll number and the name of
bubble.
the examination centre and sign with pen in the 6. The marking scheme given at the beginning of
space provided for this purpose on the original. each section gives details of how darkened and
Do not write any of these details anywhere not darkened bubbles are evaluated.
else. Darken the appropriate bubble under each 7. Zero marks ‘0’ If none of the options is chosen
digit of your roll number. (i.e. the question is unanswered).

C. FOR INTEGER TYPE QUESTIONS OMR LOOKS D. FOR DECIMAL TYPE QUESTIONS OMR LOOKS
LIKE: LIKE:
1. For example, if answer ‘SINGLE DIGIT’ integer 1.
type below : COLUMN
0 1 2 3 4 5 6 7 8 9 1 2 . 3 4
2. For example, if answer ‘SINGLE DIGIT’ integer 0 0 0 0
with positive / negative type below : 1 1 1 1
0 2 2 2 2
– 1
3 3 3 3
2
4 4 4 4
3
5 5 5 5
4
6 6 6 6
5
7 7 7 7
6
8 8 8 8
7
9 9 9 9
8
9 2. If answer is 3.7, then fill 3 in either 1st or 2nd
3. For example, if answer ‘DOUBLE DIGIT’ integer column and 7 in 3rd or 4th column.
type below :
0 0 0 0 0 0 0 0 0 3. If answer is 3.07 then fill 3 in 1st or 2nd column ‘0’
1 1 1 1 1 1 1 1 1 in 3rd column and 7 in 4th column.
2 2 2 2 2 2 2 2 2 If answer is, 23 then fill 2 & 3 in 1st and 2nd
3 3 3 3 3 3 3 3 3 column respectively, while you can either leave
4 4 4 4 4 4 4 4 4 column 3 & 4 or fill ‘0’ in either of them.
5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9

TYPE WISE CODES FOR QUESTIONS

SR# QUESTION TYPE CODE
1 MULTIPLE CHOICE QUESTION (ONLY ONE CORRECT OPTION) MCQ
2 MULTIPLE SELECT (ONE OR MORE THAN ONE CORRECT OPTIONS QUESTION) MSQ
3 NUMERICAL VALUE QUESTION NVQ
4 COMPREHENSION / PARAGRAPH BASED QUESTION CBQ
5 MATCH THE LISTING QUESTION MLQ
6 COLUMN MATCH QUESTION CMQ
7 FILL IN THE BLANKS TYPE QUESTION FBQ
8 TRUE OR FALSE TYPE QUESTION TFQ
9 ASSERTION & REASON / STATEMENT TYPE QUESTION ARQ
10 SUBJECTIVE TYPE QUESTION STQ
 MATHEMATICS
PART-I (Hkkx-I) : MATHEMATICS (xf.kr)
SECTION-1: (Maximum Marks: 12)
• This section contains THREE (03) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY if (all) the correct option(s) is(are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : − 2 In all other cases.
• For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 mark;
choosing ONLY (B) will get +1 mark;
choosing ONLY (D) will get +1 mark;
choosing no option(s) (i.e. the question is unanswered) will get 0 marks and
choosing any other option(s) will get − 2 marks.
[kaM 1 : (vf/kdre vad : 12)
 bl [kaM esa rhu (03) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA
 izR;sd iz 'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA
 vkaf'kd vad % +2 ;fnSpace
rhu ;kforrhu
Rough Work
ls vf/kd / (dPps
fodYi lghdk;Z ds fy,
gS ijUrq y nks) fodYiksa dks pquk x;k gS vkSj
dsoLFkku
nksauks pqus gq, fodYi lgh fodYi gSaA
 vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj
pquk gqvk fodYi lgh fodYi gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –2 vU; lHkh ifjfLFkfr;ksa esaA
 mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc %
 dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’
 dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) pquus ij +1 vad feysaxs ’
 dsoy fodYi (B) pquus ij +1 vad feysaxs ’
 dsoy fodYi (D) pquus ij +1 vad feysaxs ’
 dksbZ Hkh fodYi u pquus ij ¼vFkkZ r~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj
 vU; fdlh fodYiksa ds la ;kstu dks pquus ij –2 vad feysaxs

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
1. Three horses A, B and C are running a race. The event “A beats B” is denoted AB. The event A beats B
2 1
who beats C is denoted by ABC. The event “A win the race” is denoted by A. If P(AB) = , P(BC)  ,
3 2
2
P( AC)  and P( ABC)  P( ACB) , P(BAC)  P(BCA ) , P(CAB)  P(CBA ) , then
3
5 2
(A) P( A )  (B) P(B) 
9 9
3
(C) P(B)  (D) The event AB, AC, CB are pairwise independent
9
rhu ?kksMs+ A, B vkSj C ,d nkSM+ esa nkSM+rs gSA ?kVuk A,B dks gjkrk gS] dks AB ls O;Dr fd;k tkrk gS] rc ?kVuk
A, B dks gjkrk gSA tks C dks Hkh gjkrk gS] dks ABC ls O;Dr fd;k tkrk gS A ?kVuk A, nkSM+ thrrk gS] dks A ls O;Dr
2 P(BC)  1 P( AC)  2
fd;k tkrk gSA ;fn P(AB) = , , vkSj P( ABC)  P( ACB) , P(BAC)  P(BCA ) ,
3 2 3
P(CAB)  P(CBA ) , rc
5 2
(A) P( A )  (B) P(B) 
9 9
3
(C) P(B)  (D) ?kVuk AB, AC, CB Lora=k gS A
9

x
2. For every real number x  –1, let f(x) = . Write f1(x) = f(x) and for n ≥ 2, fn(x) = f(fn–1(x)). Then,
x 1
f 1(–2) · f2(–2) · · · · · fn(–2) can't be equal to
izR;sd okLrfod x  –1 ds fy, ekuk f(x) = x . ;fn f 1(x) = f(x) rFkk fn(x) = f(fn–1(x))  n ≥ 2 rc
x 1
f 1(–2) · f2(–2) · · · · · fn(–2) dk eku ugha gks ldrk &
2n 1  2n   2n 
(A) (B) 1 (C)   (D)  
1·3·5······2n  1 2  n  n

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
4
1 11
3. Let a1 < a2 < a3 < a4 be positive integers such that a
i 1 i

6
Then, a4 – a2 can't be equal to

4
1 11
ekuk a1 < a2 < a3 < a4 /kukRed iw.kkZad bl izdkj gS fd a
i 1 i

6
rc a4 – a2 dk eku ugha gks ldrk gS&
(A) 11 (B) 10 (C) 9 (D) 8.

SECTION-2: (Maximum Marks: 12)

• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : − 1 In all other cases.
[kaM 2 : (vf/kdre vad : 12)
 bl [kaM esa pkj (04) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 izR;sd iz 'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +3 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

1 1  an1
4. Let a0 = and an be defined inductively by an  ,n  1 then lim 4n 1  an  is equal to
2 2 n 

1 1  a n1
ekuk a0 = rFkk an dks vkxeukRed :i es an  ,n  1 ls ifjHkkf"kr fd;k tkrk gS rc
2 2
lim 4n 1  an  dk eku gS &
n 

2  2for Rough Work / (dPps

Space  2 dk;Z ds fy, LFkku) 2
(A) (B) (C) (D)
8 18 9 16
Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
5. In how many ways can we choose a1 < a2 < a3 < a4 from the set {1, 2,..., 30} such that a1, a2, aз, a4 are
in arithmetic progression ?
;fn leqPp; a1, a2 , a3 , a4 {1, 2,..., 30} bl izdkj gS fd a1 < a2 < a3 < a4 rFkk a1, a2, aз, a4 lekUrj Js.kh esa
gS rc budks pquus ds rjhdksa dh la[;k gS &
(A) 135 (B) 145 (C) 155 (D) 165

1 n
6. Let f : R R be a continuous function such that f(x+ 1) =
2
f(x) for all x  R, and let an =
 f x dx for
0
all integers n  1. Then
1
(A) lim an exists and equals
n  f x dx
0
(B) lim an does not exist.
n
1
(C) lim an exists if and only if
n  f x dx  1
0
1
(D) lim an exists and equals 2
n  f x dx.
0
1
ekuk f : R R ,d lrr~ Qyu bl izdkj gS fd f(x+ 1) = f(x)  x  R, rFkk izR;sd iw .kkZad n  1 ds fy,
2
n
an =
 f x dx rc &
0
1
(A) lim an fo|eku gS rFkk
n  f x dx ds cjkcj gSA
0
(B) lim an fo|eku ugha gS A
n 
1
(C) lim an fo|eku gksxk ;fn vkSj dsoy ;fn
n  f x dx  1 .
0
1
(D) lim an fo|eku gS rFkk 2
n  f x dx ds cjkcj gSA
0

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
2n 2n
 1 
 r  
r 
7. If lim
r 1  r 1
 , then =
n
n 3
4 r
r 1
2n 2n
 1 
 r  
r 
;fn nlim r 1  r 1
 rc =
n
 3
4 r
r 1

(A) 2 (B) 4 (C) 6 (D) 8

SECTION-3: (Maximum Marks: 24)

• This section contains SIX (06) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY the correct integer value is entered;
Zero Marks : 0 In all other cases.
[kaM 3 : (vf/kdre vad : 24)
 bl [kaM esa N% (06) iz'u gSaA
 izR;sd iz'u dk mÙkj xSj&_.kkRed iw.kkZad (NON-NEGATIVE INTEGER) gSaA
 izR;sd iz 'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

8. If number of functions f: {1,2,..., 10} {1,..., 2000}, which satisfy f(i + 1) – f(i)  20, for all 1 ≤ i ≤ 9,
is nCr then the least value of (n + r) is
ekuk Qyu f bl izdkj gS fd f: {1,2,..., 10} {1,..., 2000} rFkk f(i + 1) – f(i)  20,  1 ≤ i ≤ 9, ;fn ,sls
Qyuksa dh la[;k nCr gS rc (n + r) dk U;wure eku gS &

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
9. The number of consecutive zeroes adjacent to the digit in the unit's place of 40150 is
40150 esa bZdkbZ ds vad ds lehi okys Øekxr 'kwU;ksa dh la[;k gS &

9999 1
10. The integral part of  n 2
n
equals –

9999 1
 n 2
n
ds iw.kkZad Hkkx dk eku gS &

11. Let a ≥ b ≥ c ≥ 0 be integers such that 2a + 2b – 2c = 144. Then, a + b – c equals–

ekuk iw.kkZad la[;k,s a ≥ b ≥ c ≥ 0 bl izdkj gS fd 2a + 2b – 2c = 144 rc a + b – c dk eku gS &

12. Consider all the permutations of the twenty six English letters that start with z. Number of these
permutations the number of letters between z and y is less than those between y and x is n(m!) then
minimum value of (n + m) is (where n is a natural number)
z ls izkjEHk gksus okys vaxzsth o.kZekyk ds lHkh 26 v{kjksa ls cuus okys lHkh Øep;ksa ij fopkj djsA bu Øep;ksa es
mu Øep;ksa dh la[;k ftuesa z rFkk y ds e/; vkus okys v{kjksa dh la[;k y rFkk x ds e/; vkus okys v{kjksa dh
la[;k ls de gks n(m!) gS rc (n + m) dk U;ure eku gS &
¼tgk¡ n izkd`r la[;k gS½

d2 y dy dy
13. If the solution of x + = logx, where at x = 1 the value of y = 1 and = – 1, is y = xnx – px + 3,
dx 2 dx dx
then find p.
d2 y dy dy
;fn x 2
+ = logx tgk¡ x = 1 ij y = 1 vkSj = – 1 gS] dk gy y = xnx – px + 3 rc p dk eku gS
dx dx dx

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
SECTION-4: (Maximum Marks: 12)
• This section contains FOUR (04) Matching List Sets.
• Each set has ONE Multiple Choice Question.
• Each set has TWO lists: List-I and List-II.
• List-I has Four entries (P), (Q), (R) and (S) and List-II has Five entries (1), (2), (3), (4) and (5).
• FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE of
these four options satisfies the condition asked in the Multiple Choice Question.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 ONLY if the option corresponding to the correct combination is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases.
[kaM 4 : (vf/kdre vad : 12)
 bl [kaM esa pkj (04) lwph&lqesyu (Matching List) lsVl (sets) gSA
 çR;s d lwph&leqPp; (set) esa ,d ,dkf/kd fodYi ç'u (Multiple Choice Question) gSA
 çR;s d lwph&lqesyu lsV esa nks lwfp;k¡ gS : lwph-I rFkk lwph -II
 lwph-I esa pkj çfof"V;k¡ (P), (Q), (R) rFkk (S) gS rFkk lwph-II esa ik¡p (05) çfof"V;k¡ (1), (2), (3), (4) rFkk (5) gSA
 çR;s d ,dkf/kd fodYi ç'u esa lwph-I vkSj lwph-II ij vk/kkfjr pkj fodYi fn, x, gS vkSj bu fodYiksa eas ls dsoy ,d
fodYi gh ,dkf/kd fodYi ç'u dh 'krZ iwjk djrk gSA
 çR;s d ç'u ds mÙkj dk ewY;kadu fuEufyf[kr vadu ;kstuk ds vuqlkj fd;k tk,xk :
 iw.kZ vad : +3 dsoy ;fn lgh la ;kstu ds fodYi ds la xr dks pquk x;k gS A
 ‘'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ ç'u vuqÙkfjr gSA)
 _.k vad : –1 vU; lHkh ifjfLFkr;ksa esa

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
14. Match List I with List II and select the correct answer using the code given below the lists :

List-I List-II
3
(P) If (a, a2) falls inside the acute angle made by the lines (1)
2
x   
y= , x > 0 and y = 3x, x > 0, then a   , 2µ  ,
2  25 
then the value of µ is

(Q) In a ABC, if A is (0, 0), internal angle bisector through

29
vertex B is x + 2y – 5 = 0 and perpendicular bisector of side (2)
3
AC is 3x – y – 10 = 0 and if the slope of line BC is m then value
of |m| is

(R) Circles of radii 5,5 and 8 are mutually externally tangent . (3) 14
If a fourth circle of radius r touches all the three circles
externally then the value of r is :

4
(S) Consider a family of circles passing through two fixed (4)
9
points A(3, 7) and B(6, 5). If the common chords of the circle
x2 + y2 – 4x – 6y – 3 = 0 and the members of the family of
circles pass through a fixed point (a, b), then value of a + b is
8
(5)
9

(A) (P) 1 , (Q) 3,(R) 5,(S) 2 (B) (P) 3 , (Q) 1,(R) 4,(S) 2
(C) (P) 3 , (Q) 1,(R) 5,(S) 4 (D) (P) 3 , (Q) 1,(R) 5,(S) 2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s %

lwph-I lwph-II
x 3
(P) ;fn (a, a2) js[kkvksa y = , x > 0 vkSj y = 3x, x > 0 ls cus (1)
2 2
  
U;wudks.k ds vUnj gS rc a   , 2µ  fLFkr gS, rc µ dk eku gS&
 25 
(Q) ABC esa, ;fn A(0, 0) gS, 'kh"kksZ B ls xqtjus okyk vkUrfjd dks.k

v)Zd x + 2y – 5 = 0 gS vkSj Hkqtk AC dk yEc v)Zd 3x – y – 10 = 0 gS

rFkk js[kk BC dh izo.krk m gS] rc |m| Kkr dhft,A

(R) f=kT;k 5,5 ds 8 ds rhu o`Ùk] ,d nwljs dks ckg;~ Li'kZ djrs gSA (3) 14
;fn r f=kT;k dk pkSFkk o`Ùk bu lHkh rhu o`Ùkksa ds chp esa Li'kZ djrk gS]
rc r dk eku gS&

4
(S) ekukfd nks fLFkj fcUnqvksa A(3, 7) vkSj B(6, 5) ls xqtjus okys o`Ùkksa dk (4)
9
fudk; gSA ;fn o`Ùk x2 + y2 – 4x – 6y – 3 = 0 dh mHk;fu"B thok vkSj
o`Ùkksa dk fudk; ds lnL;] ,d fLFkj fcUnq (a, b) ls xqtjrk gS] rc a + b
dk eku gS

8
(5)
9

(A) (P) 1 , (Q) 3,(R) 5,(S) 2 (B) (P) 3 , (Q) 1,(R) 4,(S) 2
(C) (P) 3 , (Q) 1,(R) 5,(S) 4 (D) (P) 3 , (Q) 1,(R) 5,(S) 2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
15. Match List I with List II and select the correct answer using the code given below the lists :

List-I List-II
(P) A line through the origin intersects the parabola (1) 7
5y = 2x2 – 9 x + 10 at two points. Sum of x-coordinates
of these to points is 77. Then the slope of the line is

(Q) If P be a point on the parabola y2 = 3(2x–3) and M (2) 2

is foot of perpendicular drawn from P on the directrix
of the parabola, then length of each side of an equilateral
triangle SMP, where S is focus of the parabola is

x2 y2
(R) The value of 'a' for the ellipse   1 (a > b) , (3) 24
a2 b2
if the extremities of the latus rectum of the ellipse having
positive ordinate lies on the parabola x2 = – 2 (y – 2) is

x2 y2
(S) If the chord x cos   y sin   p , of the hyperbola   1, (4) 29
16 18
subtends a right angle at centre, and the diameter of circle,
concentric with the hyperbola, to which the given chord is a tangent,
is ‘d’, then the value of 'd' is
(5) 6
(A) (P) 2, (Q) 3,(R) 4,(S) 5 (B) (P) 5, (Q) 3,(R) 4,(S) 2
(C) (P) 1, (Q) 3,(R) 4,(S) 5 (D) (P) 2, (Q) 1,(R) 4,(S) 5

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s %

lwph-I lwph-II

(P) ewy fcUnq ls tkus okyh js[kk] ijoy; 5y = 2x2 – 9x + 10 (1) 7

dks nks fcUnqvksa ij izfrPNsn djrh gSA bu fcUnqvksa ds
x-funsZ'kkad dk ;ksx 77 gS] rc js[kk dh iz o.krk gS&

(Q) ;fn P ijoy; y2 = 3(2x–3) ij ,d fcUnq gS rFkk M, P ls (2) 2

ijoy; dh fu;rk ij yEcikn gSA rc leckgq f=kHkqt SMP dh
izR;sd Hkqtk dh yEckbZ gS tgk¡ S, ijoy; dh ukfHk gS –

x2 y2
(R) nh?kZo`Ùk   1 (a > b) ds fy, a dk eku gksxk (3) 24
a2 b2
;fn nh?kZo`Ùk ds ukfHkyEc ds fljs, ijoy; x2 = – 2 (y – 2) ij
/kukRed dksfV;ka j[krk gS&

x2 y2
(S) ;fn vfrijoy;  1 dh thok x cos   y sin   p dsUnz ij (4) 29
16 18
ledks.k cukrh gS rFkk o`Ùk dk O;kl, vfrijoy; ds lkFk ladsUnzh; gS
tks fd nh xbZ thok Li'kZ js[kk gS] d gS] rc d dk eku gS&
(5) 6

(A) (P) 2, (Q) 3,(R) 4,(S) 5 (B) (P) 5, (Q) 3,(R) 4,(S) 2
(C) (P) 1, (Q) 3,(R) 4,(S) 5 (D) (P) 2, (Q) 1,(R) 4,(S) 5

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
16. Match List I with List II and select the correct answer using the code given below the lists :

List – I List – II
2
1 x 1
(P) Number of solutions of e |x|   (1) 2
2 x2  4
(Q) sin 5° + sin 10° + …… + sin 355° = (2) 0

(R) Number of integers in the range of cos2x – sinx (3) 3

(S) Absolute value of greatest integral roots of the equation (4) 5

(x + 2)(x + 3)(x + 8)(x + 12) = 4x2 are
(5) 4

lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s %
lwph - I lwph- II
1 x2  1
(P) lehdj.k e |x|   2 ds gyksa dh la[;k gS& (1) 2
2 x 4
(Q) sin 5° + sin 10° + …… + sin 355° = (2) 0

(R) cos2x – sinx ds ifjlj esa iw.kkZadksa dh la[;k gSA (3) 3

(S) lehdj.k (x + 2)(x + 3)(x + 8)(x + 12) = 4x2 (4) 5

ds iw.kk±d ewyksa dh la[;k gS&
(5) 4

(A) (P) 3, (Q) 2,(R) 1,(S) 5 (B) (P) 1, (Q) 2,(R) 5,(S) 3
(C) (P) 1, (Q) 2,(R) 3,(S) 5 (D) (P) 4, (Q) 2,(R) 3,(S) 1

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 MATHEMATICS
17. Match List I with List II and select the correct answer using the code given below the lists :
List-I List-II
1  cos 2 x
(P) lim 2 equals (1) 1
x 0 x
e  ex  x
1/ x
 (3 / x)  1 
(Q) If the value of lim   can be expressed as ep/q, where (2) 3
x  0  ( 3 / x)  1 

p and q are relatively prime,then (p+q) is equal to

2( 3 1  x 2  4 1  2x )
(R) lim is equal to (3) 4
x 0 x  x2
x  2 sin x
(S) lim equals (4) 5
x 0
(x  2 sin x  1)  (sin2 x  x  1)
2

(5) 2

lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s %
lwph - I lwph - II
1  cos 2 x
(P) lim 2
cjkcj gS (1) 1
x0 e x  ex  x
1/ x
 (3 / x)  1 
(Q) ;fn lim   dk eku dks ep/q ds :i esa O;Dr fd;k (2) 3
x  0  ( 3 / x)  1 

tkrk gS tgk¡ p vkSj q lg vHkkT; gS rc (p+q) dk eku cjkcj gS -

3 2 4
2( 1  x  1  2x )
(R) lim cjkcj gS - (3) 4
x 0 x  x2
x  2 sin x
(S) lim cjkcj gS - (4) 5
x0
(x 2  2 sin x  1)  (sin 2 x  x  1)
(5) 2

(A) (P) 1, (Q) 2,(R) 5,(S) 4 (B) (P) 3, (Q) 4,(R) 2,(S) 1
(C) (P) 3, (Q) 4,(R) 1,(S)
Space 5 Work /(D)
for Rough (P) dk;Z
(dPps 2,ds(Q)
fy,4,(R)
LFkku) 3,(S) 1

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
PART-II (Hkkx-II) : PHYSICS (HkkSfrd foKku)
SECTION-1: (Maximum Marks: 12)
• This section contains THREE (03) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY if (all) the correct option(s) is(are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −2 In all other cases.
• For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 mark;
choosing ONLY (B) will get +1 mark;
choosing ONLY (D) will get +1 mark;
choosing no option(s) (i.e. the question is unanswered) will get 0 marks and
choosing any other option(s) will get −2 marks.
[kaM 1 : (vf/kdre vad : 12)
 bl [kaM esa rhu (03) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA
 izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA
 vkaf'kd vad % +2 ;fnSpace
rhu ;kforrhu
Rough Work
ls vf/kd / (dPps
fodYi lghdk;Z ds fy,
gS ijUrq dsoLFkku
y nks) fodYiksa dks pquk x;k gS vkSj
nksauks pqus gq, fodYi lgh fodYi gSaA
 vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj
pquk gqvk fodYi lgh fodYi gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –2 vU; lHkh ifjfLFkfr;ksa esaA
 mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc %
 dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’
 dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) pquus ij +1 vad feysaxs ’
 dsoy fodYi (B) pquus ij +1 vad feysaxs ’
 dsoy fodYi (D) pquus ij +1 vad feysaxs ’
 dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj
 vU; fdlh fodYiksa ds la;kstu dks pquus ij –2 vad feysaxs

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 PHYSICS
18. In the photo electric effect experiment, initially the wavelength of incident light is  and its intensity is .

In the second case, the wavelength of incident light is and its intensity is '. The graph of i v/s V for
2
both the case is shown in the figure. The work function () of emitter plate, threshold wavelength of the
plate th and intensity ' will be :-
çdk'k oS|qr çHkko ds ç;ksx esa çkjEHk esa vkifrr çdk'k dk rjaxnS/;Z  rFkk rhozrk gSA nqljs ç;ksx esa vkifrr

çdk'k dk rjaxnS/;Z rFkk rhozrk 'gSA nksuksa ç;ksxksa ds fy, i v/s V dk xzkQ fp=kkuqlkj gSA mRltZd IysV dk
2
dk;ZQyu (),mRltZd IysV dk nsgyh rjaxnS/;Z th vkSj rhozrk ' gksxh&
i

/2, '

, 
v
–12volt –5 volt
(A) = 2 eV (B) th  620nm (C) ' =  (D) ' = 2

19. The equivalent impudence of the circuit for very small frequency (  0) and for a very high frequency
( ) will be :
cgqr vYi vko`Ùkh (  0) vkSj cgqr mPp vko`Ùkh ( ) ij ifjiFk dh rqY; izfrck/kk gksxh %

R1 C1 L2

C2

L1 R2 R3
wt

~
Space for Rough V=V
Work / (dPps
sin(t) dk;Z ds fy, LFkku)
0

R1 R2
(A) For   0, |Z| = R2 + R3 (B) For   0, |Z| =
R1  R2
(C) For   , |Z| = R2 + R3 (D) For   , |Z| = R1 + R3

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
20. The radius of the sun is R, the radius of the earth is r E and the earth is situated at a distance ‘d’ from the
sun. The spectral distribution of the radiation emitted by the sun is shown in the figure. Treating the sun
as perfectly black body and emissivity of the earth is e, choose the correct alternative(s). (Stefan’s
constant = , Wien’s constant = b , speed of light = c)
 4R2  b 4
(A) The rate of mass loss by the sun is   
 c 2  m 
 
2 4
R  b 
(B) the intensity of sun rays falling on the earth (called solar constant S) is S =     
 d   m 
2 4
R  b 
(C) The rate of radiation energy falling on the earth is      ( r2E)
 d   m 
2 4
R  b 
(D) The rate of radiation energy absorbed by the earth is      ( r2E) (e)
 d   m 

E


m
lw;Z ls mRlftZr fofdj.kksa dk LisDVªeh ÅtkZ forj.k oØ fp=k esa iznf'kZr gSA lw;Z dh f=kT;k R, i`Foh dh f=kT;k rE
rFkk lw;Z ls i`Foh dh nwjh d gSA lw;Z dks ,d vkn'kZ d`".k oLrq ekurs gS rFkk i`Foh dh emissivity e ekurs gS A lgh
fodYi@fodYiksa dk p;u dhft,A (LVhQu fu;rkad = , ohu fu;rkad = b , izdk'k dh pky = c)
4

4R  b 
2 
(A) lw;Z esas nzO;eku gkfu dh nj =  2  
 c  m 
2 4
R  b 
(B) i`Foh ij fxjus okyh lw;Z dh fdj.kksa dh rhozrk ¼lkSj fu;rkad½ S =     
 d   m 
2 4
R  b 
(C) i`Foh ij fofdj.k ÅtkZSpace
fxjus dh
fornj =   Work
Rough   / (dPps
 (dk;Z
r2E) ds fy, LFkku)
 d   m 
2 4
R  b 
(D) i`Foh }kjk fofdj.k vo'kks"k.k dh nj =     ( r2E) (e)
 d   m 

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
SECTION-2: (Maximum Marks: 12)
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.
[kaM 2 : (vf/kdre vad : 12)
 bl [kaM esa pkj (04) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +3 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
21. The current through the circuit just after switching on, will be:
fLop can (on) djus ds rqjUr ckn ( t  0 ) ifjiFk esa cgus okyh /kkjk Kkr djks :
R L C

~
V = V0 cos(t)

 1 
 L  
(A)
V0
(B)
V0
cos  where tgkWa   tan1 C 
2 Space for Rough Work / (dPps dk;Z ds2 fy, LFkku)  R 
 1   1   
R 2   L   R2   L    
 C   C 
V
(C) 0 (D) Zero 'kwU;
R

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
22. The electron of H-atom is initially in its ground state. Monochromatic light is incident on it, in which
energy of one photon is 13.06 eV, due to which the electron transits in some excited state. If time taken
in the transition is 10–8 second, the average torque acting on the electron will be approximately:
gkbMªkstu ijek.kq dk bysDVªkWu izkjEHk esa viuh ewy voLFkk esa gSaA ,d o.khZ; izdk'k bl ij vkifrr gS ftlesa ,d
QkWVksu dh ÅtkZ 13.06 eV, ftlds dkj.k bysDVªkWu fdlh mRrsftr voLFkk esa laØfer gks tkrk gSA ;fn laØe.k esa
fy;k x;k le; 10–8 lsd.M gS rks bysDVªkWu ij dk;Zjr vkSlr cyk?kw.kZ yxHkx gksxk:
(A) 4.25 × 10–26 N.m (B) 8.25 × 10–26 N.m (C) 4.25 × 10–30 N.m (D) 8.25 × 10–30 N.m

23. How many alpha and beta particles are emitted when Uranium 92U238 decays to lead 82Pb206 ?
(A) 3 alpha particles are 5 beta particles (B) 6 alpha particles and 4 beta particles
(C) 4 alpha particles and 5 beta particles (D) 8 alpha particles and 6 beta particles
tc ;wjsfu;e 92U238 fo?kfVr gksdj ysM 82Pb206 curk gS rks fdrus ,YQk () ,oa chVk () d.k mRlftZr gksaxs ?
(A) 3  ,oa 5 d.k (B) 6  ,oa 4 d.k
(C) 4  ,oa 5 d.k (D) 8  ,oa 6 d.k

24. Two long straight cylindrical conductors with resistivities 1 and 2 respectively are joined together as
shown in figure. If current  flows through the conductors, the magnitude of the total free charge at the
interface (junction surface) of the two conductors is :
leku vuqizLFk dkV {ks=kQy ds nks yEcs /kkfRod Bksl csyu dks fp=kkuqlkj tksM+k x;k gS] ftudh izfrjks/kdrk Øe'k%
1 vkSj 2 gSA ;fn izR;sd csyu ls  /kkjk cg jgh gks rks junction lrg ij tek gksus okyk vkos'k gksxk&

 
1 2

 1  2  0
(A) zero 'kwU; (B) (C) 0 |1 – 2| (D) 0 |1 + 2|
2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
SECTION-3: (Maximum Marks: 24)
• This section contains SIX (06) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on--
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY the correct integer value is entered;
Zero Marks : 0 In all other cases.
[kaM 3 : (vf/kdre vad : 24)
 bl [kaM esa N% (06) iz'u gSaA
 izR;sd iz'u dk mÙkj xSj&_.kkRed iw.kkZad (NON-NEGATIVE INTEGER) gSaA
 izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

25. The self inductance of each coil is 4 H and their mutual inductance is 1 H when they are connected as
shown in the figure. Find the equivalent inductance of the system in Henery.
izR;sd dq.Myh dk Lo & izsjdRo 4 H gS rFkk bUgsa fp=kkuqlkj tksM+us ij budk vkil es vU;ksU; izsjdRo 1 H gS rks
fudk; dk rqY; Lo & izsjdRo Henery es Kkr djksA

26. A parallel beam of light of wavelength  is incident on a plane mirror at an angle  as shown in the

figure. If the minimum value of cos, for which a maxima occurs at the point P is (cos)min = , then
nd
write the value of n

,d lekUrj izdk'k beam rFkk rjaxnS/;Z lery niZ.k ij dks.k ij fp=kkuqlkj vkifrr gksrk gSA cos dk og
U;wure eku] ftlds fy, fcUnq
SpaceP ij
formfPp"B
RoughizWork
kIr gksr/k (dPps
gS og dk;Z ds fy, LFkku)

(cos)min = gks rks n dk eku fy[kksA
nd

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
27. Find the potential difference across the capacitor in volts in steady state
lar`Ir voLFkk esa la/kkfj=k ds fljksa ij foHkokUrj ¼oksYV esa½ gksxkA
R C

R R
R R

10V

28. In the given circuit, find the current (in mA) in the wire between points A and B.
çnf'kZr ifjiFk esa fcUnq A rFkk B ds e/; rkj esa çokfgr /kkjk dk eku (mA esa) Kkr dhft,A
1k A 2k

2k 1k
B

32V

29. A cylindrical container has cross section area of 0.01 m 2 and volume 0.1 m 3, is filled with 1 mole 'He'
gas at atmosphere pressure (P0 = 105 Pa) is covered by a massless, movable piston, A spring of spring
constant K = 100 N/m is also attached to the piston as shown. Initially the spring is unstrached and
piston is in equilibrium. Now the gas is heated very slowly, till its volume becomes double. If the
temperature of the gas becomes  times, the initial temperature, then the value of  will be :
,d csyukdkj ik=k dk vuqizLFk dkV {kS=kQy 0.01 m2 rFkk vk;ru 0.1 m3 gSA bl ik=k esa ok;qe.Myh; nkc
(P0 = 105 Pa) ij ,d eksy ghyh;e xSl Hkjh gqbZ gS rFkk bl ik=k dks nzO;ekughu xfreku fiLVu ls cUn fd;k x;k
gSA K = 100 N/m fLiazx fu;rkad dh ,d fLiazx fiLVu ls fp=kkuqlkj tqM+h gqbZ gSA izkjEHk esa fLiazx vfod`r voLFkk esa
gS] rFkk fiLVu lkE; voLFkk esa gSA vc xSl dks cgqr /khjs&/khjs rc rd Å"ek iznku dh tkrh gS] tc rd dh bldk
vk;ru nqxquk ugha gks tkrkA ;fn xSl dk rki] izkjfEHkd rki dk  xquk gSA rc  dk eku gksxk:
atmosphere
Space for Rough Work / (dPps dk;Z ds fy, LFkku)

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
30. A bus of width  and height 2 is moving on a circular road of mean radius 5m with a constant speed.
The centre of mass of the bus is at the middle of the bus, at a height  from the ground. The friction
coefficient between the tyres and the road is  = 0.6. What should be the maximum speed of the bus
(in m/s), so that it neither slides, nor topples.
 pkSM+kbZ rFkk 2 Å¡pkbZ dh ,d cl 5m ek/; f=kT;k dh o`Ùkkdkj lM+d ij fu;r pky ls xfr dj jgh gSA cl dk
nzO;eku dsUnz /kjkry ls  Å¡pkbZ ij cl ds e/; esa gS cl ds VkW;j rFkk lM+d ds e/; ?k"kZ.k xq.kkad  = 0.6 gSA
cl dh vf/kdre pky (m/s es)a D;k gksuh pkfg,] rkfd ;g u rks fQlys vkSj u gh iyVsA
V

TATA
5623

SECTION-4: (Maximum Marks: 12)

• This section contains FOUR (04) Matching List Sets.
• Each set has ONE Multiple Choice Question.
• Each set has TWO lists: List-I and List-II.
• List-I has Four entries (P), (Q), (R) and (S) and List-II has Five entries (1), (2), (3), (4) and (5).
• FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE
of these four options satisfies the condition asked in the Multiple Choice Question.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 ONLY if the option corresponding to the correct combination is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases.
[kaM 4 : (vf/kdre vad : 12)
 bl [kaM esa pkj (04) lwph&lqey s u (Matching List) lsVl (sets) gSA
 çR;sd lwph&leqPp; (set) esa ,d Space for Rough
,dkf/kd fodYi ç'u Work / (dPps Choice
(Multiple dk;Z ds Question)
fy, LFkku) gSA
 çR;sd lwph&lqesyu lsV esa nks lwfp;k¡ gS : lwph-I rFkk lwph -II
 lwph-I esa pkj çfof"V;k¡ (P), (Q), (R) rFkk (S) gS rFkk lwph-II esa ik¡p (05) çfof"V;k¡ (1), (2), (3), (4) rFkk (5) gSA
 çR;sd ,dkf/kd fodYi ç'u esa lwph-I vkSj lwph-II ij vk/kkfjr pkj fodYi fn, x, gS vkSj bu fodYiksa eas ls dsoy ,d
fodYi gh ,dkf/kd fodYi ç'u dh 'krZ iwjk djrk gSA
 çR;sd ç'u ds mÙkj dk ewY;kadu fuEufyf[kr vadu ;kstuk ds vuqlkj fd;k tk,xk :
 iw.kZ vad : +3 dsoy ;fn lgh la;kstu ds fodYi ds laxr dks pquk x;k gSA
 ‘'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ ç'u vuqÙkfjr gSA)
 _.k vad : –1 vU; lHkh ifjfLFkr;ksa esa

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
31. Match the column according to the electromagnetic waves in LIST-I with their applications given in
LIST-II, and choose the correct option.
List-I List-II
(P) micro- waves (1) TV remote, green-house effect.
(Q) infra-red light (2) Water purification in aqua-guard.
(R) Ultra-violet light (3) Taking photograph in night or foggy condition.
(S) X-ray (4) Detection of leakage of oil in underground pipeline.
RADAR system used in aircraft navigation and satellite
(5)
communication

lwph-I esa nh xbZ fo|qr pqEcdh; rjaxksa dk lwph-II esa fn, x, mi;ksxksa ds vk/kkj ij LrEHkksa dk feyku djks rFkk lgh
fodYi pqfu,A
lwph-I lwph-II
(P) micro- waves (lw{e&rjaxsa) (1) TV remote, green-house çHkko
(Q) infra-red light (vojDr&çdk'k) (2) aqua-guard esa ikuh ds 'kqf)dj.k ds fy,
Ultra-violet light (3) jkr esa rFkk dksgjs dh fLFkfr esa QksVksxzkQ ysus ds
(R) (ijkcSaxuh&çdk'k)
fy,
X-ray (4) Hkwfexr ikbiykbu esa rsy ds fjlko dk irk yxkus ds
(S)
fy,
(5) ok;q;ku ds mM+us esa ç;qDr jMkj ra=k esa rFkk
satellite communication ds fy,

Which one of the following options is correct?

dkSulk fodYi lgh :i ls lqefs yr gS?
(A) P → 5; Q → 1 ; R → 2 ; S → 4, 3 (B) P → 5; Q → 1, 3 ; R → 2 ; S → 4
(C) P → 4; Q → 2 ; R → 1, 3 ; S → 5 (D) P → 3; Q → 1, 2 ; R → 4 ; S → 5

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
32. In an alternative circuit, a bulb is used as a resistor (R). The frequency of the alternating source is
increased from   0 to    . List-I describe the change in brightness of the bulb and List-II
describe the required circuit elements.
List-I List-II
(P) If the brightness of the bulb is (1) Then the circuit should be L R series
continuously increasing circuit
(Q) If the brightness of the bulb is (2) Then the circuit should be R C series
continuously decreasing circuit
(R) If the brightness of the bulb is first (3) Then the circuit should be L C R series
increasing, and then decreasing circuit
(S) If the brightness of the bulb is remains (4) Then the circuit should contain only the
constant bulb (resistor)
(5) Not possible
Which one of the following options is correct?
(A) (P)  2, (Q)  1, (R)  3, (S)  4 (B) (P)  2, (Q)  1,(R) 4,(S) 2
(C) (P) 3, (Q) 1,(R) 5,(S) 4 (D) (P) 3 , (Q) 1,(R) 5,(S) 2
fdlh izR;korhZ lfdZV esa izfrjks/k ds :i esa ,d cYc dk iz;ksx fd;k tkr gSA izR;korhZ Lkzksr dh vko`fÙk   0 ls
   rd c<+kbZ tkrh gSA lwph-I esa cYc dh ped esa ifjorZu n'kkZ;k tkrk gS rFkk lwph -II esa vko';d ifjiFk ds
vo;oksa dks crk;k x;k gSA
lwph-I lwph-II
(P) ;fn cYc dh ped yxkrkj c<+ jgh gks (1) rks ifjiFk L R Js.khØe ifjiFk gksxkA
(Q) ;fn cYc dh ped yxkrkj ?kV jgh gks (2) rks ifjiFk RC Js.khØe ifjiFk gksxkA
(R) ;fn cYc dh ped igys c<+ jgh gks vkSj (3) rks ifjiFk L C R Js.khØe ifjiFk gksxkA
ckn esa ?kV jgh gks
(S) ;fn cYc dh ped leku jgrh gks (4) rks ifjiFk esa dsoy cYc¼izfrjks/k½ gksxk
(5) lEHko ugha gSA
dkSulk fodYi lgh :i ls lqefs yr gS?
(A) (P)  2, (Q)  1, (R)  3, (S)  4 (B) (P)  2, (Q)  1,(R) 4,(S) 2
(C) (P) 3, (Q) 1,(R) 5,(S) 4 (D) (P) 3 , (Q) 1,(R) 5,(S) 2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
33. List-I describes thermodynamic processes in four different systems. List-II gives the magnitudes (either
exactly or as a close approximation) of possible changes in the internal energy of the system due to the
process.
List-I List-II
(P) 10–3 kg of water at 100° 𝐶 is converted to steam at the (1) 2 kJ
same temperature, at a pressure of 105 𝑃𝑎. The volume of
the system changes from 10–6 to 10–3 in the process.
Latent heat of water = 2250 kJ/kg.

(Q) 0.2 moles of a rigid diatomic ideal gas with volume V at (2) 7 kJ
temperature 500 K undergoes an isobaric expansion to
volume 3 V. Assume 𝑅 = 8. 0 𝐽 𝑚𝑜𝑙 –1 K–1.

(R) One mole of a monatomic ideal gas is compressed (3) 4 kJ

1 3
adiabatically from volume 𝑉 = m and pressure 2 k Pa
3
v
to volume
8
(S) Three moles of a diatomic ideal gas whose molecules can (4) 5 kJ
vibrate, is given 9 kJ of heat and undergoes isobaric
expansion.

(5) 3 kJ

Which one of the following options is correct?

(A) (P) 1, (Q) 3,(R)  2,(S)  5
(B) (P) 3, (Q) 1,(R) 4,(S) 2
(C) (P) 1, (Q)  3,(R) 5,(S)  2
(D) (P) 3, (Q) 1,(R) 5,(S) 2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
lwph I eas pkj fHkUu&fHkUu fudk;ksa esa Å"ekxfrd izØe n'kkZ;s x;s gSA lwph II esa izØe ds dkj.k fudk; dh vkUrfjd
ÅtkZ esa lEHkkfor ifjorZuksa ds ifjek.k (;k rks Bhd ;k fudVre lfUudVu esa) fn;s x;s gSA
lwph-I lwph-II
(P) 100° ij 10–3 kg ty dks 105 Pa ds nkc ij leku rki ij (1) 2 kJ
Hkki esa :ikUrfjd fd;k tkrk gSA izØe esa fudk; dk vk;ru
10–6 ls 10–3 m3 rd ifjofrZr gksrk gSA ty dh xqIr Å"ek =
2250 kJ/kg
(Q) 500 K rki ij Vvk;ru okyh 0.2 eksy n`<+ f}ijek.kqd vkn'kZ (2) 7 kJ
xSl 3 V vk;ru rd lenkh; izlkj ls xqtjrh gSA ekuk R =
8.0 J mol–1 K–1
(R) (3) 4 kJ
,d eksy ,dy ijek.kqd vkn'kZ xSl dks vkr;u V = 1 m3
3

rFkk nkc 2 k Pa ls vk;ru v rd :)ks"e :i ls laihfM+r

8
fd;k tkrk gSA
(S) rhu eksy f}ijek.kqd vkn'kZ xSl ftlds v.kq dEiUu dj ldrs (4) 5 kJ
gS] dks 9 kJ Å"ek iznku dh tkrh gS rFkk ;g lenkch; izlkj ls
xqtjrh gSA
(5) 3 kJ

fuEufyf[kr esa ls dkSulk fodYi lgh gS ?

(A) (P) 1, (Q) 3,(R)  2,(S)  5
(B) (P) 3, (Q) 1,(R) 4,(S) 2
(C) (P) 1, (Q)  3,(R) 5,(S)  2
(D) (P) 3, (Q) 1,(R) 5,(S) 2

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 PHYSICS
34. In list-I, the displacement(y) of particles as a function of position (x) and time (t) are given. Match the list
according to most suitable phenomena
List-I List-II
(P) displacement at some position is y = (1) Equation of travelling wave
sin(1000t) cos2t
(Q) y = (sin(5t) + cos (5t)).(sin(3x) – cos(3x)) (2) Equation of standing waves
(R) y = log(3x – 5t) (3) Equation of beats
(S) 1 (4) Unaccepted function for any kind of
y= practical wave or standing wave or beat.
9 x  4t – 12xt  5
2 2

lwph-I esa d.kks dk foLFkkiu (y) fLFkfr (x) rFkk le; (t) ds Qyu esa fn;k x;k gSA lcls mi;qDr ?kVuk ds vuqlkj
lwph dk feyku djksa
lwph-I lwph-II
(P) fdlh fLFkfr ij foLFkkiu y = sin(1000t) (1) izxkeh rjax dh lehdj.k
cos2t
(Q) y = (sin(5t) + cos (5t)).(sin(3x) – cos(3x)) (2) vizxkeh rjax dh lehdj.k
(R) y = log(3x – 5t) (3) foLian dh lehdj.k
(S) 1 (4) fdlh Hkh izdkj dh HkkSfrd rjax ;k vizxkeh
y=
9 x  4t – 12xt  5
2 2
rjax ;k foLian ds fy, vLohdk;Z Qyu

Which one of the following options is correct?

fuEufyf[kr esa ls dkSulk fodYi lgh gS ?
(A) (P)  2, (Q)  1, (R)  3, (S)  4
(B) (P)  2, (Q)  1,(R) 4,(S) 2
(C) (P) 3 , (Q) 1,(R) 2,(S) 4
(D) (P) 3 , (Q)  2,(R)  4,(S) 1

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
PART-III (Hkkx-III) : CHEMISTRY (jlk;u foKku)
Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,
P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,
Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
SECTION-1: (Maximum Marks: 12)
• This section contains THREE (03) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY if (all) the correct option(s) is(are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −2 In all other cases.
• For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 mark;
choosing ONLY (B) will get +1 mark;
choosing ONLY (D) will get +1 mark;
choosing no option(s) (i.e. the question is unanswered) will get 0 marks and
choosing any other option(s) will get −2 marks.
[kaM 1 : (vf/kdre vad : 12)
 bl [kaM esa rhu (03) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA
 izR;sd iz'u ds fy, vad fuEufyf[krSpaceifjfLFkfr;ks a esa Work
for Rough ls fdlh/ (dPps
,d dsdk;Z
vuqldskjfy,
fn;sLFkku
tk;s)axs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA
 vkaf'kd vad % +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gS ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj
nksauks pqus gq, fodYi lgh fodYi gSaA
 vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj
pquk gqvk fodYi lgh fodYi gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –2 vU; lHkh ifjfLFkfr;ksa esaA
 mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc %
 dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’
 dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’
 dsoy fodYi (A) pquus ij +1 vad feysaxs ’
 dsoy fodYi (B) pquus ij +1 vad feysaxs ’
 dsoy fodYi (D) pquus ij +1 vad feysaxs ’
 dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj
 vU; fdlh fodYiksa ds la;kstu dks pquus ij –2 vad feysaxs

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 CHEMISTRY
35. Which of the following statement(s) is/are false?
(A) Azurite is a sulphide ore
(B) Slag is more dense than molten metal
(C) Al can be reduced from Al2O3 by Hooper’s electrolytic reduction process
(D) Cupellation process is used for purification of Ag
fuEu esa ls dkSulk@dkSuls dFku vlR; gS@gSa \
(A) ,tqjkbV ,d lYQkbM+ v;Ld gSA
(B) /kkrqey xfyr /kkrq ls vf/kd l?ku gksrk gSA
(C) Al gwij ds oS|qrvi?kVuh; vip;u izØe ds }kjk Al2O3 ls vipf;r gks ldrk gSA
(D) [kiZj.k izØe Ag ds 'kqf)dj.k ds fy, iz;qDr gksrk gSA

36. -maltose (C12H22O11) can be hydrolysed to glucose (C6H12O6) according to the following reaction:
C12H22O11(aq) + H2O ()  2 C6H12O6 (aq)
Given:
Standard enthalpy of formation of C12H22O11 (aq) = –2238 kJ / mol
Standard enthalpy of formation of H2O () = –285 kJ / mol
Standard enthalpy of formation of C6H12O6 (aq)= –1263 kJ / mol
Time (min .) 0 50 100
Conc. of   maltose (M ) 4.0 1.0 0.25
Which of the following statement(s) is/are true?
(A) The hydrolysis of -maltose is exothermic.
(B) Heat liberated in combustion of 1.0 mol of -maltose is greater than the heat liberated in
combustion of 2.0 mol of glucose.
(C) Increasing temperature will increase the degree of hydrolysis of -maltose.
(D) The hydrolysis of -maltose follow 1st order kinetics.

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
 -ekYVkst (C12H22O11) fuEu vfHkfØ;k ds vuqlkj Xywdkst (C6H12O6) esa tyvi?kfVr gks ldrk gSA
C12H22O11(aq) + H2O ()  2 C6H12O6 (aq)
fn;k gS %
C12H22O11 (aq) ds lEHkou dh ekud ,UFkSYih = –2238 kJ / mol
H2O ()ds lEHkou dh ekud ,UFkSYih = –285 kJ / mol
C6H12O6 (aq) ds lEHkou dh ekud ,UFkSYih = –1263 kJ / mol
le; ¼feuV½ 0 50 100
ekYVkst dh lkUnzrk ¼M½ 4.0 1.0 0.25
fuEu esa ls dkSulk@dkSuls dFku lR; gS@gSa \
(A) -ekYVkst dk tyvi?kVu Å"ek{ksih gksrk gSA
(B) 1.0 eksy -ekYVkst ds ngu esa fu"dkflr Å"ek] 2.0 eksy Xywdkst ds ngu esa fu"dkflr Å"ek ls vf/kd gksrh gSA
(C) rki c<+us ij -ekYVkst ds tyvi?kVu dh ek=kk c<+sxhA
(D) -ekYVkst dk tyvi?kVu 1st dksfV cyxfrdh ds vuqlkj gksrk gSA

37. Which of the following is/are true for products of the following reaction sequence?

 (Y) 
Br2
 (Z) 
KOH
  (P)
Conc . H2SO 4
CCl4 DMSO

(X)
(A) Y is formed by electrophilic addition on X (B) Z is formed by nucleophilic substitution of Y
(C) Formation of P from Z involves rearrangement (D) P and X are identical
fuEu es ls dkSulk@dkSuls fuEu vfHkfØ;k vuqØe ds mRiknksa ds fy, lR; gS@gSa\

 (Y) 
Br2
 (Z) 
KOH
  (P)
Conc . H2SO 4
CCl4 DMSO

(X)
(A) Y, X ij bysDVªkWuLusgh ;ksxkRed }kjk curk gSA (B) Z, Y ds ukfHkdLusgh izfrLFkkiu }kjk cukrk gSA
(C) Z ls P ds fuekZ.k es iquZfoU;kl gksrk gSA (D) P rFkk X le:i gSA

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
SECTION-2: (Maximum Marks: 12)
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.
[kaM 2 : (vf/kdre vad : 12)
 bl [kaM esa pkj (04) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +3 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

38. Which of the following ion has the highest coagulation power for the solution which is obtained by the
following reaction ?
FeCl3 (excess)  NaOH  Solution
(A) Al3+ (B) SO 24 (C) I– (D) Na+
foy;u tks fuEu vfHkfØ;k ls izkIr gksrk gS] ds fy, fuEu esa ls dkSulk vk;u mPpre Ldanu 'kfDr j[krk gS\
FeCl3 ( vkf/kD;)  NaOH  foy;u
(A) Al3+ (B) SO 24 (C) I– (D) Na+

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
39. Select the correct option :
(A) Gas is more compressible, if repulsive forces dominate over attractive forces between molecule
(B) At extremely low pressure and high temperature gases behave ideally
(C) At Boyle's temperature gases behave ideally
(D) If gas is kept at T < Tc, it can never be liquefied.
lgh fodYi dk p;u dhft,&
(A) xSl vf/kd lEihfM+r gksrh gS] ;fn v.kqvksa ds e/; izfrd"kZ.k cy] vkd"kZ.k cy ij izHkkoh gksrs gSaA
(B) vR;f/kd U;wu nkc rFkk mPp rki ij xSl vkn'kZ O;ogkj djrh gSA
(C) ckW;y rki ij xSl vkn'kZ O;ogkj djrh gSA
(D) ;fn xSl dks T < Tc ij fu;r j[krs gS] rks ;g nzohÑr ugaha gks ldrh gSA

40. Observe the following reaction

CH=CH2

HBr Mg/dry ether (1) HCHO

X Y Z
(2) H2O

H2SO4/

H3O
U W (major product)
Choose the correct option.
(A) Product U gives positive test with Lucas reagent
(B) Number of Hyperconjugable hydrogen is two in product W
(C) Product Z is 3º alcohol
(D) Product X show geometrical isomerism

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
fuEu vfHkfØ;k dk izs{k.k dhft, &
CH=CH2

HBr Mg/'kq"d bZFkj (1) HCHO

X Y Z
(2) H2O

H2SO4/

H3O
U W (eq[; mRikn)
lgh fodYi dk p;u dhft, &
(A) mRikn U yqdkl vfHkdeZd ds lkFk /kukRed ijh{k.k nsrk gSA
(B) vfrla;qXeu ;ksX; gkbMªkstu dh la[;k mRikn W esa nks gSA
(C) mRikn Z 3º ,YdksgkWy gSA
(D) mRikn X T;kfefr; leko;ork n'kkZrk gSA

OH 
H /
41.  
 'X' 
CHCl3
  'Y' 
( CH3CO )2 O
 Z
KOH CH3COONa , 

Product 'Z' will be

mRikn 'Z' gksxk
O
O
(A) (B)

O O OH
(C) (D)
CH=CH–C–OH
O

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
SECTION-3: (Maximum Marks: 24)
• This section contains SIX (06) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on--
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 ONLY the correct integer value is entered;
Zero Marks : 0 In all other cases.
[kaM 3 : (vf/kdre vad : 24)
 bl [kaM esa N% (06) iz'u gSaA
 izR;sd iz'u dk mÙkj xSj&_.kkRed iw.kkZad (NON-NEGATIVE INTEGER) gSaA
 izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %
 iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA
 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

42.

Find the number of unpaired electron in the complex (C)

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
A C
laØe.k /kkrq yo.k vkf/kD; KCN jaxghu ladqy
dk uhyk foy;u dk foy;u
vEyh; ek/;e es
H2S H2S

dkyk vo{ksi vo{ksi.k ugha

ladqy (C) esa v;qfXer bysDVªkWu dh la[;k Kkr dhft, A

43. How many of the following oxides show amphoteric nature?

fuEu esa ls fdrus vkWDlkbM mHk;/kehZ izd`fr n'kkZrs gS \
V2O3, V2O5, CrO, CrO3, Cr2O3, Mn2O7, FeO, Cu2O, ZnO.

44. What volume of air (in L) is needed for the combustion of 1m 3 of a gas having the following composition
in percentage volume 2% of C2H2, 8% of CO, 35% of CH4, 50% of H2 and 5% of non combustible gas.
The air contains 20.8% (by volume) of O2 . Give your answer after dividing by 10.
1m3 xSl] tks fd izfr'kr vk;ru esa fuEu la?kVu j[krh gS] 2% C2H2, 8% CO, 35% CH4, 50% H2 rFkk 5%
vngu'khy xSl] ds ngu ds fy, ok;q dk fdruk vk;ru ( L esa) vko';d gS\ ok;q esa (vk;ru ls) 20.8% O2
lfEefyr gSA viuk mÙkj 10 ls Hkkx yxkus ds i'pkr~ nhft;sA

45. How many of the following statement(s) is/are true ?

(a) NO2 (g) turns starch iodate paper blue.
(b) NO2 (g) turns starch iodide paper blue.
(c) Phosphorous can be extracted from bone ash.
(d) When white phosphorous is treated with hot caustic alkali, phosphine is formed.
(e) When white phosphorous is treated with hot concentrated HNO 3, NO2 is formed.
(f) If (NH4)4 Cr2O7 is heated, N2O forms.
(g) If NH3 is treated with excess Cl2, NCl3 is formed.
(h) PCl3 on hydrolysis at room temperature gives H3PO4 as one of the products.
(i) BeCl2 can act as a Lewis acid.

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
fuEu esa ls fdrus dFku lR; gS@gSa \
(a) NO2 (g) LVkpZ vk;ksMsV i=k dks uhyk dj nsrh gSA
(b) NO2 (g) LVkpZ vk;ksMkbM i=k dks uhyk dj nsrh gSA
(c) QkWLQksjl vfLFk jk[k (bone ash) ls fu"df"kZr gks ldrk gSA
(d) tc lQsn QkWLQksjl dks xeZ dkWfLVd {kkj ds lkFk mipkfjr fd;k tkrk gS] rks QkWLQhu curh gSA
(e) tc lQsn QkWLQksjl dks xeZ lkUnz HNO3 ds lkFk mipkfjr fd;k tkrk gS] rks NO2 curh gSA
(f) ;fn (NH4)4 Cr2O7 dks xeZ fd;k tkrk gS] rks N2O curh gSA
(g) ;fn NH3 dks vkf/kD; Cl2 ds lkFk mipkfjr fd;k tkrk gS] rks NCl3 curh gSA
(h) dejs ds rki ij PCl3 tyvi?kVu ij ,d mRikn ds :i esa H3PO4 nsrk gSA
(i) BeCl2 yqbZl vEy ds :i esa O;ogkj dj ldrk gSA

46. How many resonating structures of are possible (including given structure ) :


dh fdruh vuquknh lajpuk,sa lEHko gSa ¼nh gqbZ lajpuk dks ekurs gq;s½ %

47. How many equivalent (x) of grignard reagent is consumed in the given conversion:
fxzU;kj vfHkdeZd ds fdrus rqY;kad (x) fn, x;s :ikUrj.k esa miHkksx ¼[kpZ½ gksrs gSa\
SH OH
OH
SO3H
(1) CH MgBr (x eq.)
H 
3

(2)H2O
SO3H
O O Cl SH
Report your answer (10 + x).
viuk mÙkj (10 + x) ds :i esa nhft,A

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
SECTION-4: (Maximum Marks: 12)
• This section contains FOUR (04) Matching List Sets.
• Each set has ONE Multiple Choice Question.
• Each set has TWO lists: List-I and List-II.
• List-I has Four entries (P), (Q), (R) and (S) and List-II has Five entries (1), (2), (3), (4) and (5).
• FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE
of these four options satisfies the condition asked in the Multiple Choice Question.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 ONLY if the option corresponding to the correct combination is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases.

[kaM 4 : (vf/kdre vad : 12)

 bl [kaM esa pkj (04) lwph&lqey s u (Matching List) lsVl (sets) gSA
 çR;sd lwph&leqPp; (set) esa ,d ,dkf/kd fodYi ç'u (Multiple Choice Question) gSA
 çR;sd lwph&lqesyu lsV esa nks lwfp;k¡ gS : lwph-I rFkk lwph -II
 lwph-I esa pkj çfof"V;k¡ (P), (Q), (R) rFkk (S) gS rFkk lwph-II esa ik¡p (05) çfof"V;k¡ (1), (2), (3), (4) rFkk (5) gSA
 çR;sd ,dkf/kd fodYi ç'u esa lwph-I vkSj lwph-II ij vk/kkfjr pkj fodYi fn, x, gS vkSj bu fodYiksa eas ls dsoy ,d
fodYi gh ,dkf/kd fodYi ç'u dh 'krZ iwjk djrk gSA
 çR;sd ç'u ds mÙkj dk ewY;kadu fuEufyf[kr vadu ;kstuk ds vuqlkj fd;k tk,xk :
 iw.kZ vad : +3 dsoy ;fn lgh la;kstu ds fodYi ds laxr dks pquk x;k gSA
 ‘'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ ç'u vuqÙkfjr gSA)
 _.k vad : –1 vU; lHkh ifjfLFkr;ksa esa

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
48. List-I List-II
lwph–I lwph–II
(P) BF3 (1) sp hybridization ¼ladj.k½
(Q) (SiH3)3 N (2) p – p back bond ¼i'p ca/k½
(R) B2H6 (3) p – d back bond ¼i'p ca/k½
(S) N3– (4) 3c – 2e bond ¼ca/k½

Code (dwV) :
P Q R S P Q R S
(A) 1 2 3 4 (B) 2 3 4 1
(C) 3 2 4 1 (D) 4 1 2 3

49. List – I List – II

(Compounds) (Characteristics)
(P) 4.1 g H2SO3 (1) If complete neutralization is considered, then
100 meq. of reactant are present.
(Q) 4.9 g H3PO4 (2) Central atom is in its highest oxidation number.
(R) 4.5 g oxalic acid (H2C2O4) (3) 200 millimoles of oxygen atoms are present.
(S) 5.3 g Na2CO3 (4) 200 ml of 0.5 N NaOH is used for complete
neutralization.
Codes :
(P) (Q) (R) (S)
(A) 1,4 2,3 1,3,4 1,2
(B) 2,3 3,4 1,2 1,3
(C) 1,2 2,4 2,3 1,3
(D) 1,3 1,3,4 1,2,3,4 1,2,4

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
lwph – I lwph – II
(;kSfxd) (y{k.k)
(P) 4.1 g H2SO3 (1) iw.kZ mnklhuhdj.k vfHkfØ;k esa 100 feyh rqY;kad vfHkdeZd
j[krk gSA
(Q) 4.9 g H3PO4 (2) dsUnzh; ijek.kq viuh mPpre vkWDlhdj.k la[;k esa gSA
(R) 4.5 g vkWDlsfyd vEy (H2C2O4) (3) vkWDlhtu ijek.kq ds 200 feyheksy mifLFkr gSaA
(S) 5.3 g Na2CO3 (4) iw.kZ mnklhuhdj.k ds fy, 0.5 N NaOH dk 200 ml iz;qDr
fd;k tkrk gSA
dwV :
(P) (Q) (R) (S)
(A) 1,4 2,3 1,3,4 1,2
(B) 2,3 3,4 1,2 1,3
(C) 1,2 2,4 2,3 1,3
(D) 1,3 1,3,4 1,2,3,4 1,2,4

50. Match the compounds in LIST-I with the observations in LIST-II, and choose the correct option.
LIST-I LIST-II
(P) Number of structural isomeric alcohols of C4H10O (1) 3
(Q) Number of structural isomeric esters of C4H8O2 (2) 5
(R) Number of structural isomers of C6H14 (3) 6
(S) Number of structural isomers of C2BrClFI (4) 4
(5) 2
Code :
(A) P → 4; Q → 4; R → 2 ; S → 1 (B) P → 1; Q → 3; R → 3; S → 2
(C) P → 2; Q → 1; R → 1; S → 4 (D) P → 1; Q → 3; R → 2; S → 4

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
lwph-I ds ;kSfxdksa dks lwph-II ds izs{k.kksa ls feyk;sa vkSj lgh fodYi dks pqusa
lwph-I lwph-II
(P) N C4H10O ds lajpukRed leko;oh ,Ydksgyks dh la[;k (1) 3
(Q) C4H8O2 ds lajpukRed leko;oh ,LVjksa dh la[;k (2) 5
(R) C6H14 ds lajpukRed leko;fo;ksa dh la[;k (3) 6
(S) C2BrClFI ds lajpukRed leko;fo;ksa dh la[;k (4) 4
(5) 2

dwV %
(A) P → 4; Q → 4; R → 2 ; S → 1 (B) P → 1; Q → 3; R → 3; S → 2
(C) P → 2; Q → 1; R → 1; S → 4 (D) P → 1; Q → 3; R → 2; S → 4

51. Match the compounds in LIST-I with the observations in LIST-II, and choose the correct option.

LIST-I LIST-II
(Reagents reacting with PhCH2COOH) (Product formed)
(P) CH3–MgBr (1) PhCH2COCl
(Q) PCl5 (2) PhCH2COOCH3
(R) NH3, followed by heating (3) CH4
(S) CH3OH in the presence of conc. H2SO4. (4) PhCH2CONH2
(5) C6H6

Code :

(A) P → 1; Q → 2; R → 3 ; S → 4 (B) P → 3; Q → 1; R → 4; S → 2
(C) P → 4; Q → 1; R → 3; S → 2 (D) P → 3; Q → 1; R → 2; S → 4

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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 CHEMISTRY
lwph-I ds ;kSfxdksa dks lwph-II ds izs{k.kksa ls feyk;sa vkSj lgh fodYi dks pqusa
lwph-I lwph-II
¼PhCH2COOH ds lkFk fØ;k djus okys vfHkdeZd½ ¼fufeZr mRikn½
(P) CH3–MgBr (1) PhCH2COCl
(Q) PCl5 (2) PhCH2COOCH3
(R) NH3, ckn esa xeZ djus ij (3) CH4
(S) lkUnz H2SO4 dh mifLFkfr esa CH3OH (4) PhCH2CONH2
(5) C6H6

dwV %
(A) P → 1; Q → 2; R → 3 ; S → 4 (B) P → 3; Q → 1; R → 4; S → 2
(C) P → 4; Q → 1; R → 3; S → 2 (D) P → 3; Q → 1; R → 2; S → 4

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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