Narayana (Adv Level) - p1 & p2 Assignment - 2

NARAYANA IIT ACADEMY
JEE (Advanced)-2023
PAPER –1
TEST DATE:
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Time Allotted: 3 Hours Maximum Marks: 180
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General Instructions:
 The test consists of total 57 questions.
 Each subject (PCM) has 19 questions.
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 This question paper contains Three Parts.
 Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
 Each Part is further divided into Three Sections: Section-A, Section-B & Section-C.
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Section – A (01 – 04, 20 – 23, 39 – 42): This section contains TWELVE (12) questions. Each question has
FOUR options. ONLY ONE of these four options is the correct answer.
Section – A (05 –10, 24 – 29, 43 – 48): This section contains EIGHTEEN (18) questions. Each question has
FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).
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Section – B (11 – 13, 30 – 32, 49 – 51): This section contains NINE (09) questions. The answer to each
question is a NON-NEGATIVE INTEGER.
Section – C (14 – 19, 33 – 38, 52 – 57): This section contains NINE (09) question stems. There are TWO
(02) questions corresponding to each question stem. The answer to each question is a NUMERICAL
A
VALUE. If the numerical value has more than two decimal places, truncate/round-off the value to TWO
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decimal places.
MARKING SCHEME
Section – A (Single Correct): Answer to each question will be evaluated according to the following marking scheme:
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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.
Section – A (One or More than One Correct): Answer to each question will be evaluated according to the following
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marking scheme:
Full Marks : +4 If only (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 and both
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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;
Section – B: Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct integer is entered;
Zero Marks : 0 In all other cases.
Section – C: Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;
Telegram - @bohring_bot
Physics PART – I
Section – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.
1. During walking, work done by fraction is:

(A) Positive
(B) Negative
(C) Zero
(D) None of these
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2. The time period of small vertical oscillations of pendulum is: (A
pendulum bob is hanging in equilibrium with the help of two identical
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elastic wires as shown in the figure.) θ θ
(A) 2 m / YA Y, A,  Y, A, 
(B) 2 m / 2YA
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m
(C)  m / YA
(D) None of these
3. AC
The average degree of freedom per molecule of a gas is 6. The gas performs 25J work, while
expanding at constant pressure. The heat absorbed by the gas is:
(A) 75J
(B) 100J
(C) 150J
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(D) 125J
4. The radioactive sources A and B have half lives of 2hr and 4hr respectively, initially contain the
same number of radioactive atoms. At the end of 2 hours, their rates of distintegration are in the
A
ratio:
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(A) 4:1
(B) 2:1
(C) 2 :1
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(D) 1:1

This section contains SIX (06) questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR
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MOER THAN ONE of these four option(s) is (are) correct answer(s).
5. A student skates up a ramp that makes an angle 30° with the

z R y
horizontal. He/she starts (as shown in the figure) at the
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
bottom of the ramp with speed v 0 and wants to turn around x
over a semicircular path xyz of radius R during which he/she
reaches a maximum height h (at point y) from the ground as h
shown in the figure. Assume that the energy loss is negligible
30°
and the force required for this turn at the highest point is
provided by his/her weight only. Then (g is the acceleration
due to gravity)
1
(A) v02  2 gh  gR
2
3
(B) v02  2 gh  gR
2
(C) The centripetal force required at points x and z is zero
(D) The centripetal force required is maximum at points x and z
6. A rod of mass m and length L, pivoted at one of its ends, is hanging

vertically. A bullet of the same mass moving at speed v strikes the rod
horizontally at a distance x from its pivoted end and gets embedded in it.
The combined system now rotates with angular speed  about the pivot. x
L
The maximum angular speed ωM is achieved for x=x M . Then:
3vx 

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(A) = V
L  3x 2
2
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12vx
(B) = 2
L  12 x 2
L
(C) xM =
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3
V
(D) M = 3
2L
7.
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In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode)
at a distance d from the cathode. The target is kept at a potential V higher than the cathode
resulting in emission of continuous and characteristic X-rays. If the filament current I is decreased
to I/2 , the potential difference V is increased to 2V, and the separation distance d is reduced
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to d/2 , then:
(A) the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-
rays will remain the same
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(B) the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain
the same
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(C) the cut-off wavelength will reduce to half, and the intensities of all the X-rays will
decrease
(D) the cut-off wavelength will become two times larger, and the intensity of all the X-rays will
decrease
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8. Two identical non-conducting solid spheres of same mass and charge are suspended in air from
a common point by two non-conducting, massless strings of same length. At equilibrium, the
angle between the strings is  . The spheres are now immersed in a dielectric liquid of density
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-3
800 kg m and dielectric constant 21. If the angle between the strings remains the same after the
immersion, then
(A) electric force between the spheres remains unchanged
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(B) electric force between the spheres reduces

(C) mass density of the spheres is 840 kg m -3
(D) the tension in the strings holding the spheres remains unchanged
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9. Starting at time t = 0 from the origin with speed 1 ms , a particle follows a two-dimensional
x2
trajectory in the x-y plane so that its coordinates are related by the equation y  . The x and y
2
components of its acceleration are denoted by a x and a y , respectively. Then:
(A) a x  1ms 2 implies that when the particle is at the origin, a y  1ms 2
(B) a x  0 implies a y  1ms 2 at all times.

(C) at t = 0, the particle’s velocity points in the x-direction
(D) ax = 0 implies that at t = 1s, the angle between the particle’s velocity and the x axis is 45°
330π
10. A source is moving in a circle given by x 2 +y 2 =R 2 with a constant speed m/s in clockwise
6
sense. A detector is at rest at (2R, 0). Find the coordinates of the source at the instant when the
detector records the same frequency as that of the source (speed of sound = 330 m/s):
 3R R 
(A)  ,  
 2 2
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 3R R 
, 
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(B) 
 2 2 
(C)  0, R 
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(D)  R, 0 
Section – B (Maximum Marks: 12)
This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER. AC
11. Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It
expands isothermally to volume 4V1. After this, the gas expands adiabatically and its volume
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becomes 32V1. The work done by the gas during isothermal and adiabatic expansion processes
Wiso 16
are Wiso and Wadia , respectively. If the ratio  ln2 , then find the value of k ?
Wadia k
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12. A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved
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1
with a speed of 2ms in front of the open end of the pipe and parallel to it, the length of the pipe
should be changed for the resonance to occur with the moving tuning fork. If the speed of sound
1
in air is 320 ms , the smallest value of the percentage change required in the length of the pipe
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k
is , find the value of k ?
8
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 r
13. A circular disc of radius R carries surface charge density   r    0  1   , where  0 is
 R
a constant and r is the distance from the center of the disc. Electric flux through a large spherical
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surface that encloses the charged disc completely is 0 . Electric flux through another spherical
0 32
surface of radius R/4 and concentric with the disc is  . If the ratio of is , then find the
 k
value of k?
Section – C (Maximum Marks: 12)

This section contains THREE (03) question stems. There are TWO (02) questions corresponding to each
question stem. The answer to each question is a NUMERICAL VALUE. If the numerical value has more
than two decimal places, truncate/round-off the value to TWO decimal places.
Question Stem for Question Nos. 14 and 15

Question Stem
The potential difference  V  between the positive and negative terminals V  volt 
2.0
of a cell changes with the current through it as shown in the graph
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O 5 iA
14. Find the internal resistance of the cell.  in ohm 
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15. Find the emf of the cell  in volts 
Question Stem AC
A tank placed on ground has water filled in it, up to a height H. A small hole
is punched in its side wall at a height h from the bottom. Water jet strikes the
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ground at a horizontal distance x . H

Hole
(Take H = 2 meter)
h
x
A
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16. For what value  in meter  of h is x maximum?
17. For two value of h, we get same range  X  . If one value of h is H/4 , what is its other value?
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 in meter 
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Question Stem
A rod is translating on a V-shaped rail as shown in the figure at an

instant. P
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θ=37°
v0 =10 m/s
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18. Find the velocity (in m/s) of point of intersection P.
19. Find the angular velocity of intersection point P with respect to observer O.
Chemistry PART – II
20. Identify reactions correctly match with their names?

(A) O O O O Perkin
Ph C H
 i  MeCOO  / MeCOOH, Ph CH CH C OH
reaction
O
(ii) H +
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(B) O O CH3 Cannizaro
OH  reaction
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2 
 CH3
(C) Friedel Crafts
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OH OH reaction
+
H
 
O
(D) O
conc.
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reaction
2H C H H-COO   CH3 OH
NaOH
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For the gas phase reaction N 2  3H 2  2NH3 the equilibrium constant is 10

5
21. atm 2 at 300K.
G for the reaction will be:
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(A) 28.72 kJ
(B) 28.72 kJ
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(C) 14.3kJ
(D) Zero
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22. In salt analysis, cations of group IIIrd can be tested by adding certain reagents. Which is the
correct combination for the above mentioned test?
(A) Bi 3 , NH 4 Cl / NH 4 OH
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(B) Fe3 , NH 4 Cl / NH 4OH

(C) Cr 3 , H 2 S / HCl
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(D) Al 3 , NH 4 OH / H 2 S
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23. In a titration, H 2 O 2 is oxidised to O 2 by MnO4 in acidic medium. If 24 ml H2O2 requires 16ml of

0.1M MnO4 , then the volume strength of H2O2 will be:
(A) 0.167
(B) 1.89
(C) 3.72
(D) 0.93

24. Out of the following, which is not correct match for radial
probability of finding the electron of 2s orbital?
(A) A  H , B  He  , C  Li 2 
(B) A  He , B  H , C  Li 2 
(C) A  Li 2  , B  He  , C  H
(D) A  Li 2 , B  H , C  He 
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25. The compound Na 2 IrCl6 reacts with triphenylphosphine in diethylieneglycol in an atmosphere of
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CO to give  IrCl  CO  PPh 3 2  , known as ‘Vaska’s compound’.
(Atomic number of Ir = 77)
Which of the following statements is correct?
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(A) The IUPAC name of the complex is carbonylchloridebis (triphenylphosphine) iridium(III)
(B) The hybridisation of the metal ion is sp3
(C) The magnetic moment (spin only) of the complex is zero
(D) The complex shows geometrical as well as ionization isomerism
26. Choose the correct statement from the following:

(A)
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The slag obtained during extraction of iron is heavy and has lower melting point than that
of metal
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(B) At temperature below 983 K(approx) CO is chief reducing agent in blast furnance.
(C) In zone refining impurities moves in the direction of heater.
(D) Electrolytic reduction of Al2 O3 is known as Hall-Heroult process.
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27. Select the correct statement.

(A) Three moles of phenylhydrazine and one mole of aldose required to produce
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phenylosazone.
(B) Glucose form gluconic acid with Br2 /HOH
(C) Glucose form Glucaric acid with HNO3
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(D) Lowering in carbon chain can be done by Kiliani Fischer method.
28. Select the CORRECT statement.

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(A) At Boyle’s temperature a real gas behaves like an ideal gas irrespective of pressure.
TBoyle 27
(B) 
TCritical 8
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(C) On increasing the temperature four times, collision frequency (Z1) becomes double at
constant volume.
(D) At high pressure Van der Waals constant ‘b’ dominates over ‘a’.
1.1eq. of SOCl 2 Br2 Br2

HO 2 C   CH 2 3  CH 2 CO 2 H P1 P2 P3
29. +KOH /red P
2. NH 3 1eq.
 ,intramolecular SN2
  P4
Which are correct?
(A) P1 is H 2 N  CO  CH 2 3  CH 2CO 2 H

(B) P2 is H 2 N   CH 2 3  CH 2 CO 2 H
(C) P3 is H 2 N   CH 2 3  CH  Br   CO2 H
(D) P4 is
N
+
CO 2
H H

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INTEGER.
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30. Xe react directly with F2 in different condition (by volume mixture) of Xe & F2 at suitable
condition
Example: - Xe + F2   XeF2
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(2 : 1)
Ration of Xe and F2 is given to make different Xenon (Fluoride)n products.
XeF2  Xe:F2  x : a  1: 0.5
XeF4  Xe:F2  x : b AC
XeF6  Xe:F2  x : c
Find the value of a  b  c when x  2
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31. Borax and kernite both are consist of Na 2 B4 O 7 xH 2 O . Find sum of H 2 O molecules in formula of
borax & kernite.
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32. When 0.2 g of acetic acid is added to 20 g of benzene, its freezing point decreases by 0.8°C.
AN
Find the % dimerisation of acetic acid in benzene, K f (Benzene) =5 K-kg mol .

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Question Stem
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A solid compound ‘G’, of formula C15H15ON, was insoluble in water, dilute HCl, of dilute NaOH. After
prolonged heating of ‘G’ with aqueous NaOH, a liquid, ‘H’, was observed floating on the surface of the
alkaline mixture. ‘H’ did not solidify upon cooling to room temperature; it was steam-distilled and
separated. Acidification of the alkaline mixture with hydrochloric acid caused precipitation of a white
solid, ‘I’. ‘I’ is a paramethyl benzene derivative.
Compound ‘H’ was soluble in dilute HCl, and reacted with benzenesulfonyl chloride and excess KOH to
give a base-insoluble solid, ‘J’.
Compound ‘I ’ , m.p. 180°C, was soluble in aqueous NaHCO3, and contained no nitrogen.
What were compound G, H, I and J?
33. The number of Carbon atoms in ’I ‘ will be
Molecular weight of 'H'

34. The will be
53.5
Question Stem
In coordination chemistry there are a variety of methods applied to find out the structure of complexes.
One method involves treating the complex with known reagents and from the nature of reaction, the
formula of the complex can be predicated. An isomer of the complex Co(en)2(H2O)Cl2Br, on reaction with
concentrated H2SO4 (dehydrating agent) it suffers loss in weight and on reaction with AgNO3 solution it
gives a white precipitate which is soluble in NH3(aq.) [Atomic weight of Co = 59, Br = 80, Cl = 36]
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35. Number of geometrical isomers of the formula of the above original complex are (including the
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complex)
36. Molecular weight of cation of the above complex.
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Question Stem
Boron compounds are not only used for therapeutic but also for diagnostic applications. A class of
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pyrrole-containing boron compounds are used as dyes to stain and image various (transparent)
organelles inside the living cells, which are ordinarily difficult to see. The B  N bonds in these
compounds are very stable even under biochemical condition, and hence the compounds do not interfere
with cellular processes. Two representative examples of such dyes are given below:
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N N N N X=F/Cl/Br
B B
A
X X X X
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 P Q
37. What is the oxidation state of boron (B) in compound P?

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38. The degree of unsaturation among the dyes (P) and (Q) which will preferentially stain fat tissues
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Mathematics PART – III

39. P is any point, O being the origin. On the circle with OP as diameter the points Q and R are on
the same side of the diameter such that POQ = QOR = , if P, Q and R be complex numbers
z1, z2, z3 respectively such that 2 3z22 =  2  3  z1 z3 then the value  is
(A) 30
(B) 45
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(C) 15
(D) 18
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2x 2 x
40. For a function f: R  R, such that f  f  = 2f(1) for all x  R and f(0) = 0, then f(x) is
 2   2 
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(A) even function
(B) odd function
(C) periodic
(D) none of these
41. ˆ b,
If a,
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ˆ cˆ are unit vectors, then the maximum value of aˆ  bˆ  cˆ 2  bˆ  cˆ  aˆ 2  cˆ  aˆ  bˆ 2 must
be
(A) 9
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(B) 12
(C) 15
(D) none of these
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42. Let A(, 0), B(, 0), C(, 0), D(, 0) and ,  are the roots of equation ax 2 + 2hx + b = 0 while , 
are the those of a1x2 + 2h1x + b1 = 0 if C and D divides AB in the ratio  : 1 and  : 1 respectively
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also ab1, hh1, a1b are in A.P., then  +  is equal to

(A) 2
(B) 1
(C) 0
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(D) none of these

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43. If x1 and x2 are positive numbers between 0 and 1, then which of the following is/are true?
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 x  x 2  sin x1  sin x 2
(A) sin  1 
 2  2
 x  x 2  tan x1  tan x 2
(B) tan  1 
 2  2
 x  x 2  log x1  log x 2
(C) log  1 
 2  2
2
 x1  x 2  x12  x 22
(D)   
 2  2
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44. Let Sn(n  1) be a sequence of sets defined by

S1 = {0}, S2 =  
3 5
, , S3 =
2 2 
8 11 14
, ,
3 3 5 
S4 = 
15 19 23 27
, ,
4 4 4 4
, 
, … then
439
(A) third element in S20 is
20
431
(B) third element in S20 is
20
(C) sum of elements in S20 is 589
(D) sum of elements in S20 is 609
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45. The lines (m  2) x + (2m  5) y = 0, (m  1) x + (m2  7) y  5 = 0 and x + y  1 = 0 are
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(A) concurrent for three values of m
(B) concurrent for one values of m
(C) concurrent for no values of m
(D) are parallel for m = 3
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x
46. Let the function f: [–1, 1]  R is defined by f(x) =  sin t  cos t dt , then
(A)
(B)
f(x) is one-one function
periodic function
AC1
(C) monotonic function

 4 2
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(D) f(x)  0, 

  
 a2 
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47. Let f(x) = (ln(a2 – 3a – 3)) |sinx| +   cos x  x  R (where [.] denote greatest integer function)
9
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be such that f(x + T) = f(x), then

(A) if T  Q, then a  {–1, 4}
(B) if T  Q, then a  (–, –1]  {4}
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 3  21 
(C) if T  Q, then a   3, 
 2 
 3  21 
(D) if T  Q, then a   3, 
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 2 
a b c d
48. Let a, b, c, d > 0 and if x =    , then for some a, b, c, d, the
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abd abc bc d ac d

value of
1
(A) x=
2
5
(B) x=
3
3
(C) x=
2
5
(D) x=
4
12

INTEGER.
2 3 16 17
49. The polynomial p(x) = 1 – x + x – x + ... + x – x can be written as a polynomial in y where
2  k 
y = x + 1, then let coefficient of y be k then  (where [.] denote greatest integer function)
 400 
is_______
50. Let the area enclosed by the curve |x – 60| + |y| = |x/4| be 4(n!) then the value of n is__________
51. Let f(n) denote the square of the sum of the digits of natural number n, where f 2(n) denote f(f(n)),
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3 f 2011  2011  f 2010  2011
f (n) denote f(f(f(n))) and so on. The value of 2013 = ___________
 2011  f 2012  2011
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f

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Question Stem AC
Let f  x   2 x  3 ,1  x  5 and for rest of the values f  x  can be obtained by using the
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relation f  5 x    f  x  x  R .
The maximum value of f  x  in 5 ,5  for

4 5
52.   2 is__
A
AN
53. The value of f  2007  , taking   5 , is___

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Question Stem
Let f  x  be a polynomial of degree 5 with leading coefficient unity, such that

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f 1  5, f  2   4, f  3  3, f  4   2 and f  5   1 , then:

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54. f  6  is equal to____
55. Sum of the roots of f  x  is equal to____
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Question Stem
Consider a triangle ABC with vertex A  2,  4  . The internal bisectors of the angles B and C are
x  y  2 and x  3 y  6 respectively. Let the two bisectors meet at I .
56. If  a, b  is incentre of the triangle ABC then  a  b  has the value equal to___
57. If  x1 , y1  and  x2 , y2  are the co-ordinates of the point B and C respectively, then the value of
 x1 x2  y1 y2  is equal to____
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AC
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A
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JEE (Advanced)-2023
PAPER –1
TEST DATE:
Y
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ANSWERS, HINTS & SOLUTIONS
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Physics PART – I
Section – A
1. C
AC
2. A
IIT
3. B
4. C
A
5. AD
Sol. By the energy conservation (ME) between bottom point and point Y
AN
1 2 1
mv0  mgh  mv12
2 2
2 2
....  i 
AY
 v1  v0  2 gh
Now at point Y the centripetal force provided by the component of mg
 mg sin 30  mv12 / R
AR
 v12  gR / 2
From (i)
gR
 v02  2 gh
N
2
At point x and z of circular path, the points are at same height but less than h. So the velocity
more than a point y
So required centripetal  mv 2 / r is more.
6. ACD
Sol. By the angular momentum conservation about the suspension point
 m 2 
mvx    mx 2  
 3 
mvx 2vx
  2  2
m / 3  mx 2
  3x
Y
d
For maximum   0
EM
dx
 XM   / 3
So the   V / 2 3
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7. AC
hv
Sol. min 
eV
1 
 min   min  new  2
AC
V 2
dN hc
IIT
I  
dt 
dN
 decreases
dt
A
AN
8. AC
Sol. The net electric force on any sphere is lesser but by Coulomb law the force due to one sphere to
another remain the same.
AY
AR
In equilibrium Tcosα/2  mg and Tsinα/2=F

N
After immersed is dielectric liquid. As given no change in angle  .

SoT cos  / 2  mg  V  g when   800 Kg / m3 and T sin  / 2  F / er
mg mg  V  g
 
F F / er
d = density of sphere
1 800
 1  d  840
21 d
9. ABCD
10. AC
Section – B
11. 9
P1 5/3 5/3
Sol.  4V1   P2  32v1 
4
P 5/3 P
P2  1 1/8   1
4 128
Y
EM
AD
P1
PV  PV 1 1 
PV  32V1  PV 3 / 4 9
Wadi  1 1 2 2  128  1 1  PV
1 1
Wiso  PV
 1
 4V1 
1 1 ln    2 PV
5 / 3 1
1 1 ln 2
AC
2/3 8
V
 1 
IIT
Wiso 2 PV ln 2 16
 1 1  ln 2
Wadio 9 / 8PV1 1 9
A
12. 5
1 c
Sol. f   f  ......... 1
AN
1 1
( 1  initial length of pipe)
AY
 V  c
 f =
 V- VT  l2
VT speed of tuning fork,l2  new length of pipe ..... 2 
AR
1   2
 2   1  VT

N
1 V
 2  1 2 5
 100   100 
1 320 8
13. 5
R
 r
 dq 
0
0  1   2 rdr
 R
Sol. 0  
0 0
R /4
 r
 dq  
0
0 1   2 r dr
 R

0 0
Y
R
 r2 
EM
 0 2  1   dr
 0 
R
 0  R/4
  r2 
 0 2   r   dr
R
AD
0 
R2 R2

 22 3  32
R

R2
32 3  64
5 AC
Section – C
IIT
14. 0.40
15. 2.00
A
AN
16. 1.00
17. 1.50
AY
18. 12.50
19. 0.00
AR
N
Section – A
20. A
21. B
Sol. G    RT ln Kp
22. B
3 3 3
Sol. Group – 3 cations  Cr , Fe , Al
Y
Reagent  NH 4 Cl/NH 4 OH
EM
23. B
Sol. Meq of KMnO4 = Meq. Of H2O2
16×0.1×5=24×x×2
16  0.1 5
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x=  0.167 M
24  2
Volume strength  11.2  Molarity  11.35  0.167  1.89
24.
Sol.
ABD
r=r0
n2
AC
Z
IIT
As Z increases, r0 decreases
Hence upon increasing Z, nuclear distance of node decreases.
25. C
A
26. BCD
Sol. The slag formed is CaSiO3 which is lighter and has less melting point.
AN
27. ABC
Sol. Kiliani Fischer method is for up gradation of C – chain.
AY
28. BCD
Sol. At Boyle’s Temperature; gas behaves ideally for a range of pressure.
AR
29. ABCD
Section – B
30. 51
N
Sol. For XeF2  Xe : F2  2 :1

For XeF4  Xe : F2  2 :10
For XeF6  Xe : F2  2 : 40
Sum  1  10  40  51
31. 14
Sol. Borax =Na 2 B4 O7 .10H 2 O
Kernite =Na 2 B4 O7 .4H 2 O
32. 8
  0.2 / 60
Sol. 0.8  1    5 
 2 0.02
  0.08
Section – C
33. 8.00
Sol. G is an amide
34. 2.00
Y
EM
35. 2.00
36. 295.00
Sol.  Co  en  2 BrCl  Cl.H 2 O
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37. 3.00
38. 13.00
Sol. AC
Nitrogen & halogens are more EN the boron. Number of H2 molecular required saturating the
compound.
IIT
A
AN
AY
AR
N

Section – A
39. C
2 2
 z2   OQ  i2 z 3 OR i2
Sol.  z    OP  e ,  e
 1   z1 OP
z 22 cos2 2  3
  
z1z3 cos2 2 3
3
Y
 cos2 = .
2
EM
  = 150.
40. B
Sol. f(x) = ax + b
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f(0) = 0  b = 0.
41. B
Sol. = 3  aˆ  aˆ  bˆ  bˆ  cˆ  cˆ   2  aˆ  bˆ  bˆ  cˆ  cˆ  aˆ 
max is 3  3 + 2 .
3
2
= 12.
AC
42. C
IIT
2h
Sol. +=–
a
b
 =
A
a
2h1
AN
+=–
a1
b1
 =
a1
AY
   
given  = =
 1 

AR
similarly  = .

                  
Now  +  =
      
N
2
 2hh1  a1b  ab1 
aa1
= = 0.
      
43. B, D
Sol. The sine curve chord going two points on the curve lies below the curve, hence (A) cannot be
true. In log curve also, the same pattern is following but in tan curve and x2 curves opposite
pattern is followed.
44. A, C
12  1 22  1 32  1
Sol. The 1st elements of S1, S2, S3, S4, … are , , , 
1 2 3
t2  1
 Si must be a set starting at
i
S20  
399 399  20 399  40
20
,
20
,
20
,    20 term 
439
 third term is
20
20  2  399 
S 20    19  1  589.
2  20
Y

EM
45. C, D
m  2 2m  5 0
2
Sol. D  m  1 m  7 5  0
AD
1 1 1
 (m  3) (m2  m + 2) = 0
For m = 3, the lines become parallel.
46. A, C, D
x
AC
Sol. f(x) =  sin t  cos t dt is an increasing function.
1
IIT
47. A, D
A
48. B, C, D
Sol. a + b + c + d > {a + b + d, a + b + c, b + c + d, a + c + d}
AN
x>1
Let max{a, b, c, d} = a
 a + b + c > b + c + d; a + c + d > b + c + d
b c d bcd
AY
   
abc bc d ac d bc d
a a
x– 1  x < 1 + 2
abd abd
AR
Hence 1 < x < 2.
Section – B
N
49. 2
1  x18
Sol. p(x) =
1 x
 1   y  118 
coefficient of y2 in   is coefficient of y3 in (1 – y)18 = 18
C3 = k = 816
 y 
 k 
  = 2.
 400 
50. 5
51. 1
Sol. f(2011) = (2 + 0 + 1 + 1)2 = 16
f 2(2011) = (1 + 6)2 = 49
f 3(2011) = (4 + 9)2 = 169
f 4(2011) = (1 + 6 + 9)2 = 256
f 5(2011) = (2 + 5 + 6)2 = 169
 f 2n(2011) = 256
f 2n + 1(2011) = 169.
Section – C
Y
52. 32.00
EM
53. 1118.00
54. 120.00
AD
55. 15.00
56. 2.00
57. 5.00 AC
IIT
A
AN
AY
AR
N
JEE (Advanced)-2023
PAPER –2
TEST DATE:
Y
Time Allotted: 3 Hours Maximum Marks: 180
EM
General Instructions:
 The test consists of total 57 questions.
 Each subject (PCM) has 19 questions.
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 This question paper contains Three Parts.
 Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
 Each Part is further divided into Three Sections: Section-A, Section-B & Section-C.
Section – A (01 –06, 20 – 25, 39 – 44): This section contains EIGHTEEN (18) questions. Each question has
AC
FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).
Section – A (07 –10, 26 – 29, 45 – 48): This section contains SIX (06) paragraphs. Based on each
paragraph, there are TWO (02) questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE
of these four options is the correct answer.
IIT
Section – B (11 – 13, 30 – 32, 49 – 51): This section contains NINE (09) questions. The answer to each
question is a NON-NEGATIVE INTEGER.
Section – C (14 – 19, 33 – 38, 52 – 57): This section contains NINE (09) question stems. There are TWO
(02) questions corresponding to each question stem. The answer to each question is a NUMERICAL
A
VALUE. If the numerical value has more than two decimal places, truncate/round-off the value to TWO
decimal places.
AN
MARKING SCHEME
Section – A (One or More than One Correct): Answer to each question will be evaluated according to the following
marking scheme:
AY
Full Marks : +4 If only (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 and both of which
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are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option;
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Section – A (Single Correct): Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen.
Section – B: Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct integer is entered;
Section – C: Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;
Physics PART – I

1. Suppose we make a frame of a regular tetrahedron using six O

I
resistors with the same resistance, r and apply a voltage V
between point O and point M as shown in the figure. A current I
flows into point O and flows out of point M. M is the midpoint of
resistor BC. C
Y
(A) The current between O and B is 3/8I M
(B) The current between A and B is I/8 I
EM
(C) The potential at point B is 2V/5 A B
(D) The equivalent resistance between point O and M is 5r/8
2. Two point monochromatic and coherent sources of light of

D
AD
wavelength l are placed as shown in figure. The initial phase
difference between the sources is 0. D >> d. Mark the
CORRECT statement(s).   O
S1 S2
AC d
Screen
7
(A) If d= , O will be minima
2
IIT
(B) If d=  , only one maxima can be observed on screen

(C) If d  4.8 , then total 10 minima would be there on screen.
5
A
(D) If d  , then intensity at O would be minimum.

2
AN
3. The two switches in the circuit shown in figure have been closed for a long time. At t = 0, switch 1
is opened. Then 35 ms later, switch 2 is opened:
4 t=0 3 t=35 ms
AY
S1  S2
12 6 18
AR
60V 150 mH
(A) The current through the inductor for t<35ms is 6e 40t A

60 t  0.035 
N
(B) The current through the inductor for t  35 ms is1.48 e A

(C) Heat lost in the 18 resistor is 845.27 mJ .
(D) Heat lost in the 3 resistance is 618.24 mJ .
4. At the center of a hollow hemisphere uniformly charged with a surface charge density  , there is
a small dipole of dipole moment  . The moment of inertia about the rotation axis is I .

(A) The magnitude ofE due to hemisphere at the center of hemisphere is  /2 0

(B) The minimum potential energy of dipole is  
4 0
(C) If dipole is released from rest from a position perpendicular to the symmetry axis, the

maximum KE of dipole is
4 0
(D) The dipole performs SHM for small angular displacements with angular frequency is
ρE
ω=
I
5. A spherical shell of radius a is just filled with liquid of density ρ and the system rotates with
uniform angular velocity ω about vertical diameter.
2
(A) The level from center at which the pressure on the shell is maximum is g/ω .
Y
2
(B) The level from center at which the pressure on the shell is maximum is 2g/ω
EM
5 1
(C) The thrust force due to liquid on the lower half of the shell is ρgπa 3 + ρπω2a 4
3 4
1 1
(D) The thrust force on upper half of the shell is ρgπa 3 + πρω2a 4
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3 4
6. Let the energy, magnitude of linear momentum and angular frequency of an electron in hydrogen
atom be E , ρ and ω respectively. If n be the corresponding quantum number then:
(A)
E
  varies as n.
AC
ω
 Eρ 
IIT
(B)   is independent of n.
 ω 
(C)  ρω varies as n1/2
A
(D)  Eρω is independent of n.

AN

This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
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answer.
Paragraph for Question Nos. 07 and 08

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In a homogeneous, non-magnetic, highly insulating and viscous medium, a moving particle experiences a
 
viscous drag given by the law f  bv . Here b is a positive constant. A particle having charge q
projected with an unknown velocity from a point in the medium almost stops (its speed becomes
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practically negligible) after travelling a distance of 10 m in a straight line. Now a uniform magnetic field is
established in the region and the same particle is again launched with the same projection velocity
directed perpendicular to the magnetic field.
7. In the presence of above magnetic field, if the particle almost stops at a point 6m away from the
point projection the magnetic field is equal to:
4b
(A)
3q
3b
(B)
4q
4q
(C)
3b
3q
(D)
4b
8. The strength of the magnetic field is now doubled in magnitude. How far (in meters) from the
point of projection will the particle come to an almost rest?
73
(A)
2
20
Y
(B)
73
EM
40
(C)
73
30
(D)
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73
AC
A rod of length 10 meter has one end on smooth floor and the other end on
smooth wall is released from the rest from the position shown in figure. When
the rod makes an angle of 37° with the horizontal, answer the following:
IIT
53
9. The angular velocity of rod is:

A
(A) 3 rad/sec
AN
(B) 5 rad/sec
5
(C) rad/sec
AY
3
3
(D) rad/sec
5
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10. The velocity of centre of mass of rod is:

(A) 5 m/s
N
(B) 10 m/s
(C) 15 m/s
(D) 20 m/s

INTEGER.
11. Two large circular discs separated by a distance of

0.01 m are connected to a battery via a switch as
shown in the figure. Charged oil drops of density 900
3
kg m are released through a tiny hole at the center
of the top disc. Once some oil drops achieve terminal
velocity, the switch is closed to apply a voltage of 200
V across the discs. As a result, an oil drop of radius
8 107 m stops moving vertically and floats between
Y
the discs. The number of electrons present in this oil
drop is ________.
EM
2
(Neglect the buoyancy force, take acceleration due to gravity =10ms and charge on an electron
(e) = 1.6 1019 C)
AD
12. A hot air balloon is carrying some passengers, and a few sandbags of mass 1 kg each so that its
total mass is 480 kg. Its effective volume giving the balloon its buoyancy is V. The balloon is
floating at an equilibrium height of 100 m. When N number of sandbags are thrown out, the
balloon rises to a new equilibrium height close to 150 m with its volume V remaining unchanged.
AC
If the variation of the density of air with height h from the ground is   h    0 e
 0  1.25 kg m 3 and h 0  6000 m , the value of N is_____.
 h / h0
, where
IIT
13. A thermally isolated cylindrical closed vessel of height 8 m is kept

vertically. It is divided into two equal parts by a diathermic (perfect thermal
conductor) frictionless partition of mass 8.3 kg. Thus the partition is held
initially at a distance of 4 m from the top, as shown in the schematic figure 8m
A
below. Each of the two parts of the vessel contains 0.1 mole of an ideal
gas at temperature 300 K. The partition is now released and moves
AN
without any gas leaking from one part of the vessel to the other. When
equilibrium is reached, the distance of the partition from the top (in m) will
be.
(Take the acceleration due to gravity =10 ms-2 and the universal gas constant = 8.3 J mol-1 K-1).
AY

AR

N
Question Stem
Two small balls have charges q1 and q2 and have same velocity. Both of them enter into a uniform
electric field simultaneously. After some time the direction of motion of the first ball changes by 60° and its
speed get reduced half. The direction of motion of the second ball changes by 90° in the same interval.
14. The speed of second ball reduced by a factor of_____
15. Find the ratio of specific charges of the two balls. Ignore any interaction between the bass.
Question Stem
Three identical cylinders of rough surfaces are mounted on frictionless horizontal A C

axles passing through their centres. They are all spinning about their axles in anti-
clockwise direction with same angular speed 0 . They are brought closer into  
0
contact and their axels are held stationary till slipping between them ceases. [The
cylinders are held symmetrically relative to one another as shown, so that all of P
them touch one another] ( ω0 =12 rad/sec ) 
Y
Final angular speed ( rad/sec ) of cylinder B will be____
EM
16.
17. The arrangement of the three cylinders is changed as shown in figure. Now, what is their final
angular velocity ( rad/sec ) of they are brought into contact?
AD
A B C
  
0 0 0
AC
Question Stem
IIT
A uniform rope of length L=2.5meter is placed over a smooth pulley and held in a position, where
40% of its length lies on one side and the remaining 60% hangs on the other side. The rope is
released from this position. [Neglect dimensions of the pulley] (g = 10 m/s2)
A

AN
AY
A
B
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18. Find the vertical acceleration  in m/s 

2
of the COM of the rope, immediately after it is
released.
N
19. 
Find the vertical velocity in m/s
2
 of the COM, at the instant end A is L/5 above its initial
position.

20. Correct statement among the following is:

(A) Ethoxymethyl chloride reacts with Nucleophiles, faster than 1-Chlorobutane
(B) Ethoxymethyl chloride may follow SN1mechanism when reacts with Nucleophiles
(C) Phenylthio-2-ethyl chloride reacts with water faster than Ethoxymethyl chloride
(D) Higher rate of 2-Phenylthioethyl chloride is because the neighbouring group participation
by phenyl ring.
Y
21. Correct acidic strength order of the given carbonyl compound is:
EM
O O O O O O O O
AD
Cl
I  II   III   IV 
(A) IV>I>II>III
(B)
(C)
(D)
IV>II>III>I
IV>III>II>I
IV>I>III>II
AC
22. Correctly matched among the following is:
IIT
Column I (reaction) Column II (Major product)

(A) O  CH3 2 C  OH  CH 2COCH 3
Ba  OH2
2 H3C C 

CH3 
 Diacetone alcohol
A
 yield is increased in soxhlet apparatus 

AN
(B) 
HCl
2CH 3COCH 3   CH 3 2 C  CHCOCH3
Mesityl oxide
AY
(C)  i  HCl
 CH3 2 C  CHCOCH  C  CH 3 2

2CH 3COCH 3  
 ii  CH COCH 
3 3
Phorone
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(D) 
CH 3CHO+CH3 COCH 3 
NaOH
CH 3CH  OH  CH 2COCH 3
4-Hydroxypentan-2-one
N
23. The cell shown below generates a potential of 0.6 V at 300 K. Find log K sp   for AgBr :
Ag AgBr  s  NaBr  0.1 M  AgNO3  0.1 M  Ag
 2.303RT 
 Take F
=0.06 

(A) 30.2
(B) 12
(C) 12
(D) 8.9
24. Correctly matched reaction among the following is:

(A) NH2 NH2
 
PhN 2 Cl
slightly acidic medium
N N Ph
 major 
(B) NH2 NH2
Y
HNO3 +H 2SO4
EM
NO 2  major 
AD
(C) NH2 NH2
Cl
H3C CH2 CH2 /anhyd. AlCl3
AC  major 
CH3
IIT
H3C
(D) NH2 NH2
Br Br
Br2 /HOH
A
AN
Br  major 
AY
25. Correct order of Rate of Reduction among the following is:

(A) O O
 R     (Reduction with Borane)
AR
R C OH C Cl
(B) O O
R C H  R C OH      Reduction with LiAlH 4 
N
(C) R-C  C-R > R-CH  CH-R      Reduction with H 2 /Ni 

(D) O O
R C Cl  R C OH      Reduction with DiBAL-H 

answer.
A metal complex having composition Cr  NH3 4 Cl 2 Br has been isolated in two forms (A) and (B). The
form (A) reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia,
whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia.
Now given the answers to following questions:
Y
26. The compound (A) and (B) are:
EM
(A)  Cr  NH3  4 BrCl  Cl and Cr  NH 3  4 Cl 2  Br
(B)  Cr  NH3  4 Cl2  Br and  Cr  NH 3  4 BrCl  Cl
AD
(C)  Cr  NH 3 3 .Cl2 Br  .NH3 and Cr  NH 3  4 BrCl  Cl
(D)  Cr  NH 3 4 .Cl2  Br and  Cr  NH 3 3 .Cl 2 Br  .NH 3
27. The magnetic moment of A and B are, respectively:

(A) 3.87 BM, 1.36 BM
AC
(B) 1.36 BM, 3.87 BM
IIT
(C) 3.87 BM, 3.87 BM

(D) 1.36 BM, 1.36 BM

A
AN
Novocaine, a local anesthetic, is a compound with molecular formula C13H 20O 2 N 2 . It is insoluble in
water and dilute NaOH, but soluble in dil. HCl. Upon treatment with NaNO2 and HCl and then with  -
naphthol, a highly coloured solid is formed.
AY
When Novocaine is boiled with aqueous NaOH , it slowly dissolves. The alkaline solution is shaken with
ether and layers are separated.
Acidification of the aqueous layer causes the precipitation of white solid ‘A’; continued addition of acid
causes ‘A’ to redissolve. Upon isolation ‘A’ is found to have melting point of 185 -186° C and the
AR
formula C7 H 7 O2 N .
Evaporation of the ether layer leaves a liquid ‘B’ of formula C6 H15ON. ‘B’dissolve in water to give a
N
solution that turns litmus blue. Treatment of ‘B’ with acetic anhydride gives 'C', C8 H17 O 2 N , which is
insoluble in water and dilute bases, but soluble in dilute HCl.
‘B’ is found to be identical to the compound formed by the action of diethylamine on ethylene oxide.
28. When compound ‘B’ is treated with p – nitrobenzoyl chloride followed by moderate reduction with
Ni/H 2 ; the compound formed is:
(A) O
NH2 C OCH2NH2
10
(B) O
NH2 C O CH2CH2N(CH2CH3)2
(C) O
NH2 COCH2NH2OH
(D)
NO 2 CH2OH
Y
NaNO 2 ,HCl β-naphthol
29. Novocaine    Product
EM
(A) O OCH2CH2N(CH2CH3)2
C
AD
(B) O
AC
N N C OCH 2CH2N(CH2CH 3) 2
IIT
OH
A
AN
(C) OH
N O
AY
N COCH2CH2N(CH2CH3) 2
(D) OH
AR
N
N N CH2 CH 2N(CH 2 CH3 )2

INTEGER.
30. An organic compound P contains 62.07% carbon and 10.34% hydrogen and rest oxygen. Its
vapour density is 29. This compound does not react with sodium metal, but its 2.9 g combines
with X g of bromine (to give dibromo addition product). Find out value of (Y–X). [Where Y is total
number of possible isomers (including tautomers also) of given organic compound P]
[Atomic mass: Br = 80]
11
31. One mole ideal monoatomic gas is heated in two different process
according to path AB and AC. If temperature of state B and state C are
q AC
equal. Calculate 10 
q AB
32. 6.84 gm Al 2  SO 4 3 is needed to coagulate 2.5 L of As 2S3 sol completely in 2.0 hrs. The
coagulation value of Al2  SO 4 3 in terms of millimoles per litre is:
[Atomic mass: Al = 27, S = 32]
Y
EM
AD
Question Stem
Treatment of 2, 4-pentanedione with KCN and acetic acid, followed by hydrolysis, gives two products, A
and B. Both A and B are dicarboxylic acids of formula C7 H12 O 6 . A melts at 98°C . When heated, B
AC
gives first a lactonic acid ( C7 H10 O5 , m.p. 90°C) and finally a dilactone ‘Q’( C7 H8 O 4 , m.p. 105°C),
(A) What structure must B have that permits ready formation of both a monolactone and a dilactone?
(B) What is the structure of A?
IIT
33. 1mole of ‘A’ reacts with _______moles of Na.
34. The degree of unsaturation in ‘Q” is

A

AN
Question Stem
An ideal gas whose adiabatic exponents equal to 1.5 is expanded according to law
1
AY
P  atm   1.2V  Litre  . The initial volume of the gas is equal to L .The final volume becomes 4 times
3
the initial volume.
AR
35. Increment in internal energy _____litre atm
36. Magnitude of the work litre atm performed by the gas______ litre atm.

N
Question Stem
The reaction of Cl 2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts to two
(different) oxoacids of chlorine, P and Q, respectively. The Cl 2 gas reacts with SO 2 gas, in presence of
charcoal, to give a product R. R reacts with white phosphours to give a compound S. On hydrolysis, S
gives an oxoacid of phosphours T.
37. The number of oxygen atoms in ‘Q’ is.
38. The number of atoms in a molecules ‘R’ is
12

2 -1
39. The values of x for which x – 4x > cot x holds true belong to
(A) (-, -1)
(B) [5,)
(C) (-1, 5]
(D) None of these
Y
2 3
40. From (1, 2), tangents are drawn to the curve y – 2x – 4y + 8 = 0. Then
EM
(A) Sum of x–coordinates of points of contact is zero
(B) Sum of x–coordinates of points of contact is 4
(C) Sum of y–coordinates of points of contact is zero
(D) Sum of y–coordinates of points of contact is 4
AD
b b x
41. If  f x dx   f  x dx a  b  also f(x)  0 for any x  (a, b) and g  x    f  x dx , then
b
a a
 f x g x dx when a and b are positive

AC 0
a
(A) Cannot be positive
IIT
(B) Cannot be negative

(C) Cannot be equal to zero
(D) Nothing can be said
A
42. If , ,   R, then the determinant

AN
(e i  e  i ) 2 (e i  e  i ) 2 4
i
 = (e  e i ) 2 (e i  e i ) 2 4 is
i  i 2 i i  2
(e  e ) (e  e ) 4
AY
(A) equal to 0
(B) equal to e i       
(C) independent of ,  and 
AR
(D) dependent of ,  and
43. If AB = A and BA = B, where A and B are square matrices, then

A2 = A
N
(A)
(B) A2 = B
(C) B2 = B
2
(D) B =A
44. If a complex number satisfies the equation |z| – log2(z + 2) = 0 (Re z > 0), then
(A) z = z
(B) |z| = 2

(C) arg(z) =
2
(D) arg(z) = 0
13

answer.
a, b, c, d is an A.P. with integer terms and d  a 2  b 2  c 2 .
45. The common difference is

(A) 1
(B) 2
Y
(C) 3
(D) none of these
EM
46. The value of a  2b  3c  4d is
(A) 0
(B) 10
AD
(C) 100
(D) none of these
f ( x )  sin 2x  {x}: x  [0,10] .

AC
47. Number of points where f achieves local maximum is
IIT
(A) 10
(B) 11
(C) 20
(D) none of these
A
48. Number of roots of f ( x )  0 in (0, 10) is

AN
(A) 20
(B) 30
(C) 31
(D) none of these
AY

AR
INTEGER.
49. f is a function such that f ( x)  2 f (1  x)  x 2  1 . Value of f (3) is

N
50. Degree of differential equation of family of ellipses of same size having minor axis along a given
line is
51. In ABC , tan A : tan B : tan C  1: 2 : 3. If sin A : sin B : sin C  5 : 2 2 :  , then  
14

Question Stem
Consider a chess board shown as
Y
EM
AD
52. The number of ways in which 1 white and 1 black square can be selected such that they are not
in same row or same column is_____
53. AC
The number of ways in which a pair of black squares be chosen such that they have exactly one
corner in common is____

IIT
Question Stem
Let log 3 N  1  1
A
log 5 N   2   2
AN
log 7 N   3   3
Where 1 ,  2 and  3 are integers and 1 ,  2 ,  3   0,1 .
AY
54. Number of integral value of N if 1  4 and  2  2 ___

AR
55. Difference of largest and smallest integral value of N if 1  5,  2  3 and  3  2 .

N
Question Stem
Let f  x  and g  x  be two differentiable functions, defined as:

f  x   x 2  xg ' 1  g ''  2  and g  x   f 1 x 2  xf '  x   f ''  x  .
56. The value of f 1  g  1 is_____
f  x
57. The number of integers in the domain of the function F  x     3  x is____
g  x
JEE (Advanced)-2023
PAPER –2
TEST DATE:
Y
EM
ANSWERS, HINTS & SOLUTIONS
AD
Physics PART – I
Section – A
1.
Sol.
ABD
By Kirchoff's rule,
AC
I  3
1 ri  r   i   r  I  2i   i  I
IIT
2  8
I O
i
A
i
I-2i
AN
I C
-i M I
2
I 2
A I
I B
AY
-i 2
2
r I 5
 2 V   ri  rI
22 8
AR
V 5r
R eq = 
I 8
N
2. ABCD
Sol. x at O = d [path difference in maximum at O]
So if d  7 / 2, O will be a minima
d   , O will be maxima
d  5 / 2, O will be minma and hence intensity is minimum
d  4.8  then total 10 minima can be observed on screen 5 above O and 5 below O, which
corresponds to
 3 5 7 9
x   , , , ,
2 2 2 2 2
3. ABCD
4. BCD

Sol. (B) U    E   
4 0
 
(C)   KE  0  KE 
4 0 4 0

ρ
Y
E
(D) Iα=- E         E / I
EM
I
5. ACD
AD
6. AB
7. A
8.
Sol.
D
(for Q. 7-8)
For the charge
AC
   

F  q v  B  bv
IIT
  
dv q  dr   b  dr 
    B  
dt m  dt  m  dt 

 q   b  r 
A
v 
m

r  B  m
AN
9. D
AY
10. C
Sol. UsingIAOR
  1
mg sin 53  mg sin 37  I 2
AR
2 2 2
3 3
  Vcm  5  15
5 5
N
Section – B
11. 6
V 200
Sol. E= = =2×104 V/m
d 0.01
When terminal velocity is achieved qE=mg
4 3
 n  1.6  1019  2  104 
3
 8  107   900  10  n  6
12. 4
Sol. 480×g=v1 g
 480  N  g  v2 g
vρg
mg
Y
480  N  2

480 1
EM
N 50 50  480
1  1 N 4
480 6000 6000
AD
13. 6
Sol. Assuming temperature remains constant at 300 K
From P1V1 =P2 V2
V 
 2  2
V 
P1  0  P1'  0  Ax 

AC
T T
IIT
A
AN
 2x 
3 2 
1
AY
 16  x 
6 x  16  x 2
x 2  6 x  16  0
AR
x2
Distance 4+2 = 6
Section – C
N
14. 0.58
15. 0.75
16. 0.00
17. 4.00
18. 0.40
19. 1.20
Section – A
20. A, B, C
Sol. Neighboring group participation by a sulpher than involved.
21. C

Sol. Process of active methylene group  acidic strength and the side chain should be e
withdrawing.
Y
22. A, B, C
Sol. Yield of 4-Hydroxypentan-2-one is low because CH3 CH(OH) CH2 CH3 will form.
EM
23. B
 
Sol. Anode Ag  Br  AgBr  e
AD
 
Cathode Ag  e  Ag
0
Ecell   ERP
0
   ERP0 
C A
0
E
Q
cell  0.06 log K sp

1


1
 100
AC
 Br   Ag  0.1 0.1
IIT
0
Ecell  0.06 log K sp  0.06 log100
0.643  0.06 log K sp  0.12
0.763
A
log K sp   12.71
0.06
AN
24. A, B, D
Sol. Friedel Craft reaction not shown by Highly activated Rings
AY
25. A, B, C
Sol. DiBAL-H is electrophilic reducing agent as Borane and readily reduces electron rich species.
26. A
AR
Sol. AgCl is white precipitate and AgBr is pale yellow:
27. C
N

Sol. No of unpaired e are 3.
28. B
29. B
Section – B
30. 1
Sol. Molecular formula from calculation comes to be C3H6O i.e., it stands for the following compounds.
OH
O
O O
O
OH
OH OH O CH2
Y
H2C CH3
Cis and trans
EM
Br
Br2
O
O
AD
Br
VD×2=MM =29×2=58
 58g of ether combines with 2×80g Br2
 2.9g of ether will combines with

2  80
58
AC
 2.9  8g of Bromine
Y=10 and X=8
Y - X = 10-8 =2
IIT
31. 8
Sol. C V =3R/2
A
AC polytropic P  V
AN
-1
 P.V =constant
R
C AC  C V   2R
1   1
AY
AB isobaric
C AB =C p  5 R / 2
AR
QAC C AC
  T same for both 
Q AB C AB
2R 4
N
   0.8
 5R  5
 
 2 
32. 8
Sol. M mole of Al2  SO 4 3  20
Section – C
33. 4.00
34. 4.00
Sol. O
O
'Q'
O
CH3 CH3
O
Y
35. 4.00
EM
36. 1.00
Sol. (for Q. 35 – 36)
1 4
Initial volume V0 = L, final volume Vf = L
AD
3 3
Initial pressure P0 =0.4 atm, final pressure Pf =1.6 atm
0.4 6.4
TInitial =
3×nR
ΔU=nCv,m  Tf -T0 
, Tfinal =
3nR AC
R  6  R  6   6 
IIT
=n×   =n×   =2nR  4

  1  3nR  1.5  1  3nR   3nR 
Work by the gas = Pext dv

A
=  1.2V dv
AN
1.2  2
=  4V0  -V0 2 
2  
2
1.2×15V02 1.2×15  1 
AY
=      1litre atm
2 2 3
37. 3.00
AR
38. 5.00
Sol. Cl 2 +SO 2 
 SO 2 Cl2
N
R 
10SO 2 Cl2  P4 
 4PCl5  10SO 2
R  S
PCl5 +4H 2 O 
 H 3 PO 4  5HCl
S T
Section – A
39. A, B
40. B, D
Sol. Curve is (y – 2)2 = 2x3 – 4 …(1)
2
dy 3x

dx y  2
Y
 Equation of tangent at (x1, y1)
3x12
EM
y  y1  x  x1 
y1  2
As it passes through (1, 2)
3x12
AD
 2  y1  1  x1 
y1  2
or  y1  2 2  3 x12  x1  1
or
or
2 x13  4  3x13  3 x12 [As (x1, y1) lies on the curve]
x13  3 x12  4  0
AC
or (x1 + 1) (x1 – 2)2 = 0
 x1 = –1 or 2
IIT
But x1  = –1 (it is clear from equation (1))

 x1 = 2
 y1 = 2  2 3
A
Points of contact are (2, 2 + 2 3 ) and (2, 2 – 2 3 )

AN
 sum of x–coordinates of points of contact is 4

 sum of y–coordinates of points of contact is 4.
41. B, C
AY
b b
Sol. If  f x dx   f x dx
a a
AR
 f(x) is either positive in (a, b)

or f(x) is negative in (a, b) and also b > a

Also g  x  f x 
N
When f(x) > 0  g (x) > 0  g(x) is increasing function

When f(x) < 0  g (x) < 0  g(x) is decreasing function
also g(0) = 0
b b
 f  x g  x dx   g  x g  x dx
a a
=
g b  2
  g a 
2
which will be positive for 0 < a < b.
2
42. A, C
Sol. C1  C1  C2
4 (e i  e i ) 2 4
i  i 2
= 4 (e  e ) 4 =0 ( C1 & C3 are identical).
4 (e i  e i ) 2 4
43. A, C
Sol. We have AB = A
 A (BA) = A [ BA  B ]
 (AB) A = A
Y

 AA = A  AB  A
 A2 = A
EM
Again BA = B
 B (AB) = B  AB  A
 (BA) B = B
AD
 BB = B  B2 = B
44. B, D
Sol. |z| – log2(z + 2) = 0
z + 2 = 2|z|
x + 2 = 2x and y = 0
hence x = 2, y = 0.
AC
IIT
45. A
46. B
A
47. A
AN
48. D
Section – B
AY
49. 0
50. 1
AR
51. 3
Section – C
N
52. 768.00
53. 49.00
54. 44.00
55. 99.00
56. 3.00
57. 2.00

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NARAYANA IIT ACADEMY

Full Marks : +3 If ONLY the correct option is chosen.

of which are correct;

1. During walking, work done by fraction is:

Section – A (Maximum Marks: 24)

MOER THAN ONE of these four option(s) is (are) correct answer(s).

5. A student skates up a ramp that makes an angle 30° with the

6. A rod of mass m and length L, pivoted at one of its ends, is hanging

(B) electric force between the spheres reduces

(B) a x  0 implies a y  1ms 2 at all times.

Section – C (Maximum Marks: 12)

Question Stem for Question Nos. 14 and 15

14. Find the internal resistance of the cell.  in ohm 

Question Stem for Question Nos. 16 and 17

ground at a horizontal distance x . H

16. For what value  in meter  of h is x maximum?

A rod is translating on a V-shaped rail as shown in the figure at an

18. Find the velocity (in m/s) of point of intersection P.

20. Identify reactions correctly match with their names?

(C) Friedel Crafts

For the gas phase reaction N 2  3H 2  2NH3 the equilibrium constant is 10

(B) Fe3 , NH 4 Cl / NH 4OH

Section – A (Maximum Marks: 24)

26. Choose the correct statement from the following:

27. Select the correct statement.

(D) Lowering in carbon chain can be done by Kiliani Fischer method.

28. Select the CORRECT statement.

1.1eq. of SOCl 2 Br2 Br2

(A) P1 is H 2 N  CO  CH 2 3  CH 2CO 2 H

Section – B (Maximum Marks: 12)

Find the % dimerisation of acetic acid in benzene, K f (Benzene) =5 K-kg mol .

Section – C (Maximum Marks: 12)

Question Stem for Question Nos. 33 and 34

33. The number of Carbon atoms in ’I ‘ will be

Molecular weight of 'H'

Question Stem for Question Nos. 35 and 36

36. Molecular weight of cation of the above complex.

37. What is the oxidation state of boron (B) in compound P?

Mathematics PART – III

also ab1, hh1, a1b are in A.P., then  +  is equal to

(D) none of these

Section – A (Maximum Marks: 24)

44. Let Sn(n  1) be a sequence of sets defined by

(C) monotonic function

(D) f(x)  0, 

be such that f(x + T) = f(x), then

abd abc bc d ac d

Section – B (Maximum Marks: 12)

Section – C (Maximum Marks: 12)

Question Stem for Question Nos. 52 and 53

The maximum value of f  x  in 5 ,5  for

53. The value of f  2007  , taking   5 , is___

Question Stem for Question Nos. 54 and 55

Let f  x  be a polynomial of degree 5 with leading coefficient unity, such that

f 1  5, f  2   4, f  3  3, f  4   2 and f  5   1 , then:

54. f  6  is equal to____

55. Sum of the roots of f  x  is equal to____

Question Stem for Question Nos. 56 and 57

In equilibrium Tcosα/2  mg and Tsinα/2=F