Reshuffling_Test-1_JEEA.pdf

FIITJEE Ltd., Bhubaneswar Centre, Bharat Scouts & Guides Complex, Scouts Bhawan, 2nd Floor, C-IX/8, Unit – 3, BBSR-22, Odisha. 1
RESHUFFLING TEST-1
PHYSICS, CHEMISTRY & MATHEMATICS
Pattern - RT-1
Time Allotted: 3 Hours Maximum Marks: 183
 Please read the instructions carefully. You are allotted 5 minutes specifically for
this purpose.
 You are not allowed to leave the Examination Hall before the end of the test.
INSTRUCTIONS
Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet
before attempting the paper. Wrong CODE or no CODE will give wrong results.
A. General Instructions
1.Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2.This question paper contains Three Sections.
3.Section-I is Physics, Section-II is Chemistry and Section-III is Mathematics.
4.Each Section is further divided into Two Parts: Part-A & B.
5.Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided for
rough work.
6.Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any
form, are not allowed.
B. Filling of OMR Sheet
1.Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR sheet.
2.On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment No. and
write in ink your Name, Test Centre and other details at the designated places.
3.OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Two Parts.
(i) Part-A (01-07) – Contains seven (07) multiple choice questions which have One or More correct answer.
Full Marks: +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks: +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option
is darkened.
Zero Marks: 0 If none of the bubbles is darkened.
Negative Marks: 1 In all other cases.
For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in
+4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in 1
marks, as a wrong option is also darkened.
(i) Part-A (08-13) – Contains six (06) multiple choice questions which have ONLY ONE CORRECT answer Each
question carries +3 marks for correct answer and -1 marks for wrong answer.
(ii) Part-B (01-05) contains five (05) Numerical based questions, the answer of which maybe positive or negative numbers
or decimals (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) and each question carries +3 marks for correct answer and
there will be no negative marking.
Name of the Candidate :____________________________________________
Batch :____________________ Date of Examination :___________________
Enrolment Number :_______________________________________________
BATCH
–
TWO
YEAR
CRP(2123) FIITJEE – Reshuffling Test (JEE-Advanced)
QP Code: PAPER - 1

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PART – A (One or More Than One Correct Type)
This section contains 07 multiple choice questions. Each question has 4 choices (A), (B), (C)
and (D), out of which ONE or MORE THAN ONE is correct.
1. A uniform disc of mass m and radius R is kept on frictionless horizontal table. Two particles of mass
m are connected to disc by two identical light inextensible threads a shown in figure. The particles are
given velocity v0 perpendicular to the length of strings. Then
(A)acceleration of the particle will be
2
0
10
v
R
(B) angular acceleration of disc will be
2
0
2
2
5
v
R
(C) tension in the string is
2
0
20
mv
R
(D)tension in the string is
2
0
10
mv
R
2. AB is a boundary divides two media I and II having bulk modulus in the ratio 3 : 4 and density in the
ratio 4 : 3. Speed of sound in medium I is 300 m/s. Observer can move with a maximum speed of 300
m/s whereas source is stationary. The original frequency is n.
(A)for heating maximum frequency observer have to move with speed
(B) for hearing maximum frequency observer have to move at an angle 1 3
tan
4
  
 
 
with x-axis
(C) for hearing maximum frequency observer have to move at an angle 1 4
tan
3
  
 
 
with x-axis
(D)maximum frequency heard by the observer is 2n
3. Figure (Five straight lines numbered 1, 2, 3, 4 and 5) shows graph of change in temperature T versus
heat supplied Q for different processes performed on a gas of one mole.

(A)line 4 and 5 (coinciding with Q and T axis respectively) represent isothermal and adiabatic
processes respectively
(B) line 2 represents isochoric process for diatomic gas
(C) line 1 and 3 represent isobaric process for a diatomic gas and isochoric process for a monoatomic
gas respectively
(D)line 2 represents isobaric process for a monoatomic gas
4. A train is moving with constant acceleration a along positive x-axis. A particle is projected inside the
train in a vertical plane containing the motion of the train as shown. Which of the following can be the
trajectory of the particle be as observed from inside the train?
(A) (B)
(C) (D)
5. Two light wires A and B shown in the figure are made of the same material and have radii rA and rB
respectively. The block between them has a mass m. When the force F is
3
mg
, one of the wires breaks.
(A)A breaks if rA = rB
(B) A breaks if rA < rB
(C) Either A or B may break if rA = 2rB
(D)The lengths of A and B must be known to predict which wire will break
6. A tunnel is dug along a chord of the earth at a perpendicular distance
2
R
from the earth’s centre. The
wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel.
The pressing force by the particle on the wall and the acceleration of the particle varies with x
(distance of the particle from the centre) according to

(A) (B)
(C) (D)
7. A solid block of mass 2 kg is resting inside a cube as shown in the figure. The cube is moving with a
velocity ˆ ˆ
5 2
v ti j
 
r
m/s. Here, t is time in second. The block is at rest with respect to the cube and
coefficient of friction between the surfaces of cube and block is 0.6. Then, [take g = 10 m/s2
]
(A)force of friction acting on the block is 10 N
(B) force of friction acting on the block is 4 N
(C) the total force exerted by the block on the cube is 14 N
(D)the total force exerted by the block on the cube is 10 5N
(Single Correct Choice Type)
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out
of which ONLY ONE is correct.
8. A cubical block ‘A’ of mass m1 and edge length ‘a’ lies on a smooth horizontal floor. It has a groove
with a flat base with an open end. On the closed end of the groove there is a spring on natural length
‘a’ attached to it. A small block ‘B’ of mass m2 is pushed into the groove compressing the spring by
2
a
. Coefficient of friction between B and groove is . System is now released from rest. Find speed of
B as it comes out of A. Assume that B is always sliding against the surface of the groove. (m1 = 3 kg,
m2 = 1 kg, a = 10 cm, k = 100 N/m,  = 0.13)
(A)15 cm/s (B) 10 cm/s (C) 5 cm/s (D)20 cm/s

9. The end of capillary tube is immersed into a liquid. Liquid slowly rises in the tube up to a height. The
capillary fluid system
(A)will absorb heat (B) will released heat
(C) will not be involved in any heat transfer (D)nothing can be said
10. Some successive frequencies of an ideal organ pipe are measured as 150 Hz, 250 Hz, 350 Hz. Velocity
of sound is 340 m/s.
(A)it must be an open organ pipe of length 3.4 m
(B) it must be an open organ pipe of length 6.8 m
(C) it must be an closed organ pipe of length 1.7 m
(D)it must be an closed organ pipe of length 3.4 m
11. In the situation as shown in figure time period of vertical oscillation of block for small displacements
will be
(A) 2 cos
2
m
k
  (B) 2 sec
2
m
k
  (C) 2 sin
2
m
k
  (D) 2 cosec
2
m
k
 
12. A small cars engine can deliver 90 kw of power. The cars mass is 1000 kg. How fast can the car move
on a level road if the total resistance force is proportional to velocity i.e. friction
F v

 where  is 100
Ns/m. If the speed of car is
(A)10 m/s (B) 20 m/s (C) 30 m/s (D)None of these
13. Three identical rods AB, CD and PQ are joined as shown. P and Q are mid-points of AB and CD
respectively. ends A, B, C and D are maintained at 0°C, 100°C, 30°C and 60°C respectively. the
direction of heat flow in PQ is
(A)from P to Q (B) from Q to P
(C) heat does not flow in PQ (D)data not sufficient
Part – B Numerical based questions
This section contains 5 questions. The answer to each question may be positive or negative numbers or
decimals (e.g. 6.25, 7.00, -0.33, 30.27, -127.30).
1. In the figure shown there is no slipping anywhere. Mass of plank and each sphere is m. The ratio of
acceleration of C.M. of bigger and smaller sphere will be ______________

2. Two particles A and B of same mass are connected by means of light inextensible string of length l and
kept on an inclined plane of inclination
3
sin
5
 
 

 
 
. The coefficient of friction between A and plane
is  where as no friction act between B and inclined plane. B is projected with some speed to complete
circle on inclined plane. The minimum value of  for completion of circle is
21
n
, find the value of n.
3. Friction coefficient between any two surfaces in contact is 0.5. Pulleys and strings are frictionless.
Find the friction between blocks A and B (in Newton)
4. A uniform solid sphere of mass m = 400 gm and radius R = 2 cm is released from rest from a point
A of a rough slide AB. Initially, the centre O of the sphere is at the horizontal level of A. at the lower
end B, the slide passes to smooth horizontal plane. A spring is attached to a wall on the horizontal
plane. Find the maximum compression (in cm) of the spring in the process of motion of the sphere.
(Take g = 10 m/s²)
5. A syringe is filled with water upto volume 20 cm3
. The area of cross-section of the cylinder is 5 cm2
.
The syringe is held vertically and its 90 gm piston is pushed upward by external agent with constant
speed. A water beam coming out of the small nozzle (hole area 1 mm2
) has speed 2 m/s. Neglecting
friction and viscous nature of water find the work done by the agent [in 10–2
J] in fully emptying the
syringe. (Take g = 10 m/s2
)

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This section contains 07 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONE or MORE THAN ONE is correct.
1. A gas obeys the equation P(V-b) = RT. Which of the following is/are correct about the graphs of gas?
(A)The isochoric curves have slope =
R
V b

(B) The isobaric curves have slope =
R
P
and intercept b
(C) For the gas compressibility factor = 1+
Pb
RT
(D)For the gas compressibility factor = 1
Pb
RT
2. Select correct statement(s) for the reaction:
H2O(g) + CO(g)  H2(g) + CO2(g)
Substance CO(g) CO2(g) H2O(g) H2(g)
o
f
H
 , 400 K(K cal mol–1
) – 25 – 95 – 55 0
o
400K
S (cal mol–1
K–1
) 45 50 40 30
(A)Reaction is enthalpy driven (B) reaction is entropy driven
(C) Reaction is spontaneous at 400 K (D)reaction is non-spontaneous at 400 K
3. 1 2
3
4
5
6
Which of the following statement(s) is/are correct about the above molecule?
(A)C1 – C2 and C3 – C6 have same bond order
(B) 1 2 3 6 2 3
C C C C C C
     (decreasing bond order)
(C) 1 2 4 5
C C and C C
  have same bond length
(D) 1 2
C C
 bond has more double bond character than that of 1 2
C C
 of 1, 3-Butadiene
4. Consider energy graph for the AES. The true statement(s) is/are
E1

(A) E1 for nitration on
F
is greater than on
Br
(B) E1 for nitration on is greater than on toluene
(C) E1 for Friedel Craft methylation on
NH2
is greater than benzene
(D) E1 for Fiedel Craft methylation on
D D
D
D
D
D
is greater than benzene
5. Which of the following quantum numbers is/are not allowed?
(A) n 3, 2,m 0
  
l (B) n 2, 2,m 1
   
l (C) n 3, 0, m 1
  
l (D) n 5, 2, m 1
   
l
6.
COOH COOH
COOH COOK
| & | behave as acid as well as reducing agent. Then which of the following are the
correct statements regarding
COOH
COOH
and
COOK
COOH
(A)When behaves as reducing agent, then its equivalent weights are equal to half of its molecular
weight respectively
(B) 1000 mL of 1N solution of each is neutralized by 1000 mL 1N Ca(OH)2
(C) 1000 mL of 1M solution of each is neutralized by 1000 mL of 1M Ca(OH)2
.
(D)1000 mL of 1M solution of each is neutralized by 200 mL 2M of KMnO4
in acidic medium.
7. Among the following which compound is / are having lower pKa value than benzoic acid?
(A)HCO2H (B) Picric acid (2, 4, 6-trinitro phenol)
(C) CH3CO2H (D)O-nitrophenol
8. Which of the following order is correct?
(A) 3 3 3
PH NH AsH
  (Boiling point)
(B) 3 3 3
NF NCl NBr
  (Basicity)
(C) 2 2 2
NO NO N O


  (Bond angle)
(D) 2
3 2
CO CO CO

  (C – O bond length)

FIITJEE Ltd., Bhubaneswar Centre, Bharat Scouts & Guides Complex, Scouts Bhawan, 2nd Floor, C-IX/8, Unit – 3, BBSR-22, Odisha.
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9. What is the major product expected from the following reaction?
(A) (B)
(C) (D)
10. Two buffers, (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA.
Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal
volumes of the two buffers? (KHA
= 1.0 × 10-5
).
(A)4.7033 (B) 5.7033 (C) 6.7033 (D)8.7033
11. 0 0
200 120
4 2
.2
Strongly heated
C C
Heat
B CaSO H O A
C

 


A, B and C are respectively:
(A)plaster of Paris, dead burnt plaster, calcium sulphide
(B) dead burnt plaster, plaster of Paris, lime
(C) plaster of Paris, dead burnt plaster, calcium sulphite
(D)plaster of Paris, dead burnt plaster, calcium oxide (lime)
12. An inorganic compound (A) is white solid and exists as dimer in the vapour state. A shows the
following properties.
Compound (A) get sublimes on heating
Compound (A) gives fumes (B) with wet air
Compound (A) gives white ppt with NH4OH. However (A) is soluble in excess of NaOH to give
soluble compound (C). Then the inorganic compound ‘A’ is
(A)Al2O3 (B) BCl3 (C) AlCl3 (D)BeCl2
13. In the given figure bulb contains an excess of volatile solid A. At 0
25 C equilibrium formed as shown
below
At equilibrium; total pressure inside bulb is 10 mm whereas partial pressure of 2
H S inside bulb is 4
mm. If some A(g) is removed from bulb at equilibrium
(A) decomposition equilibria will shift forward
(B) sublimation equilibria will shift forward

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(C) total pressure when equilibrium reestablished will be equal to 16 mm.
(D) decomposition equilibria will shift backward direction
decimals (e.g. 6.25, 7.00, -0.33, 30.27, -127.30).
1. Total possible stereoisomers of following compound
3 3
CH CHCl CH CH CHCl CH
    
2. Two moles of ideal monoatomic gas were taken through isochoric heating from 100 K to 800 K. If the
process is carried out irreversibly (one step), then system surrounding
S S
   is (in cal/K) ____________
[Given: ln 2 = 0.7 and R = 2 cal/mol K]
3. How many moles of KMnO4
are needed to completely oxidise a mixture of 1 mole of each FeSO4
&
FeC2
O4
in acidic medium?
4. A mixture of 2 gases – A (reactive) and X(inert) is kept in a closed flask when A decomposes as per
the following reaction at 500 K: (decomposition follows 1st
order kinetics)
A(g)  2B(g) + 3C(l) + 4D(s)
P° = 210 mm Hg
Pt (10 mins) = 330 mm Hg
P = 430 mm Hg
Vapour pressure of C(l) at 500 K is 20 mm Hg.
Calculate the half-life for A (in mins).
5. Each of following reactions are carried out and the products are separated by careful fractional
distillation (OR) recrystallization. For each reaction find out how many fractions will be collected?
(a) 1-chloropentane + Cl2 (300°C)  C5H10Cl2. Number of fractions collected are ‘x’.
(b) (S)-2-chloropentane + Cl2 (300°C)  C5H10Cl2. Number of fractions collected are ‘y’.
(c) (S)-3-chloro-1-butene + HCl  2,3-dichloro-2-methylbutane. Number of fractions collected are ‘z’.
Then x + y + z is ___________

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This section contains 07 multiple choice questions. Each question has 4 choices (A), (B), (C)
and (D), out of which ONE or MORE THAN ONE is correct.
1. In ABC if ac = 3, bc = 4 and cos (A – B) =
3
4
then
(A) measure of angle A is
2

(B) measure of angle B is
2

(C)
C
cot 7
2
 (D) circum-radius of ABC is 1/2
2
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2. The first term of an infinite geometric series is 21. The second term and the sum of the series are both
positive integers. Then which of the following is are true?
(A) The second term cannot be a multiple of 5
(B) The sum of the possible values of the second term is 64
(C) The minimum value of the third term is
48
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(D) The third term can take only one integral value
3. z1, z2, z3 are three non-zero distinct points satisfying |z – 1| = 1 & 2
2 1 3
z z z
 . Then
(A) 3 2
2 3
z z
z z 2

 
is purely imaginary (B) 3
2
1 2
z
z 1
arg 2arg
z 1 z
   


   

   
(C) 3
2
1 1
z
z 1
arg 2arg
z 1 z
   


   

   
(D)
2 3 1 2 1 3
1 1 1 1 1 1
z z z z z z
    
4. Three distinct normals are drawn to the parabola y² = 4x from point S lying on the line 4x + 5y = 0. If
P, Q, R be the feet of co-normal points such that PR subtends right angle at Q, then coordinates of S
is/are
(A) (15, –12) (B) (20, –16) (C) (10, –8) (D) (5, –4)
5. If the equation 1 + p sin² x = p² – sin4
x, p  0 has a solution, then ‘p’ CANNOT be equal to
(A)
2
5
 (B)
11
3
(C)
5
4
(D)
12
5
6. Four points z1, z2, z3, z4 in complex plane such that 1
3
z
5
 , |z2| = 1, |z3|  1 and
 
2 1 4
3
1 4
z z z
z
z z 1



then
|z4| can be equal to
(A) 7
5
log (B) 2
3
log (C) e
log (D) e
log
7. If the roots of the quadratic equation  
2
x 2 a 1 x a 5 0
     are such that at least one root lies in the
interval (1, 3) then the parameter ‘a’ can lie in the interval
(A)
3
, 1
2
 
 
 
 
(B)
6
, 1
5
 
 
 
 
(C)
4 8
,
3 7
 
 
 
 
(D)
8 1
,
7 2
 
 
 
 

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8. Let A be a point on the ellipse
 2 2
x 2 y
1
16 12

  with B & C as its foci. Then the locus of the incentre
of ABC is
(A) an ellipse with eccentricity
1
2
(B) an ellipse with eccentricity
2
3
(C) a hyperbola with eccentricity
3
2
(D) an ellipse with eccentricity
3
2
9. Person A selects 4 distinct numbers from the set (1, 2, 3, 4, 5, 6, 7, 8, 9) and arranges them in
descending order to form a 4 digit number. Person B selects 4 distinct numbers from set (1, 2, 3, 4, 5,
6, 7, 8) and also arranges them in descending order to form a 4 digit number. Then the no. of ways
where person A’s 4-digit number is greater than person B’s 4-digit number is
(A)  
8 8 8
3 4 4
C C 1 C
  (B)  
8 8 8
3 4 4
1
2 C 2 C 1 C
2
   
(C)  
8 8 8
3 4 4
C 2 C 1 C
   (D)  
8 8 8
3 4 4
1
2 C C 1 C
2
  
10. Let PQR be a triangle with largest angle P. Also, the incircle of PQR touches PQ, QR, RP at points
A, B, C respectively such that PA, QB, RC are consecutive odd integers. If the radius of circumcircle
of PAC is 4 5 units, then perimeter of triangle PQR is
(A)104 (B) 110 (C) 102 (D)100
11. If |ax² + bx + c|  2  x  [–1, 1] then the maximum value of the expression
|cx² + 2bx + 4a|  x  [–2, 2] is __________.
(A)22 (B) 14 (C) 16 (D)20
12. The number of ways of arranging 9 boys (B1, B2, …., B9) and 5 girls linearly such that B1 and B2 are
not together and exactly four girls are together is equal to
(A) 280 5 9 (B) 370 4 9 (C) 15210 5 (D) 24010 5
13. If a tangent to the hyperbola
   
2 2
2 1
1
16 25
y x
 
  with slope
1
2
 touches it at the point (m, n) then
m + n equals
(A)
5
1
39
 (B)
7
1
39
 (C)
11
1
39
 (D)
8
1
39

decimals (e.g. 6.25, 7.00, -0.33, 30.27, -127.30).
1. Let ‘z’ be a non-real complex number satisfying the equation z50
= 1. If
49
17k 34k
k 1
1
N
1 z z


 
 then N =
__________.
2. If 1 1
1 x
tan 5tan x
1 x
 

 

 

 
then the values of x satisfying the above equation is/are ______________.

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3. The number of factors of 217
38
512
which are either perfect squares or perfect cubes (but not both) is
equal to __________.
4. Consider the ellipse
2 2
x y
1
16 9
  . Let P, Q, R, S be four point on this ellipse such that the normal
drawn from these points are concurrent at (2, 5). Then the centre of the conic on which these 4 points
lie is (, ) where  –  is equal to __________
5. The value of    
14
k 3 17
k
k 0
1 15 k C

 
 is equal to __________.

15
RESHUFFLING TEST-1
PHYSICS, CHEMISTRY & MATHEMATICS
ANSWERS
Sl.No. Physics Sl.No. Chemistry Sl.No. Mathematics
1. ABD 1. ABC 1. BC
2. ABD 2. AC 2. BC
3. ABCD 3. BCD 3. ABD
4. ABC 4. BC 4. AC
5. ABC 5. BC 5. ABD
6. BC 6. ABD 6. BC
7. AD 7. AB 7. BC
8. B 8. D 8. B
9. B 9. A 9. D
10. C 10. B 10. C
11. B 11. D 11. C
12. C 12. C 12. B
13. A 13. B 13. B
1. 1 1. 6 1. 33
2. 4 2. 7.35 2. 1
3. 0 3. 0.8 3. 369
4. 4 4. 10 4. 11
5. 8 5. 14 5. 25
FIITJEE – Reshuffling Test (JEE-Advanced)

16
CONCEPT CODE
Sl.No. Physics Sl.No. Chemistry Sl.No. Mathematics
1. P110703 1. C111204 1. M111502
2. P111307 2. C111905 2. M110502
3. P111205 3. C111301 3. M110307
4. P110308 4. C111804 4. M110906
5. P111101 5. C110106 5. M110101
6. P110902 6. C111103 6. M111302, M111304
7. P110404 7. C111302 7. M110106
8. P110602 8. C110305 8. M111001
9. P111011 9. C111707 9. M111206
10. P111306 10. C110504 10. M111504, M111506
11. P111104 11. C110904 11. M110103
12. P110506 12. C111508 12. M111204
13. P111207 13. C110402 13. M111103
1. P110710 1. C111404 1. M110305
2. P110507 2. C111906 2. M121401
3. P110406 3. C111101 3. M111203
4. P110710 4. C110604 4. M111006
5. P110503 5. C111703 5. M110404

17
HINTS & SOLUTION
PHYSICS:
1. (ABD)
2 2
0
2
2 2
v mR
T m R and TR
R
 
 
   
 
 
2. (ABD)
B
v

 1 300 /
v m s
 , 2 400 /
v m s

Use refraction of sound wave
1 2
2 2
3
4
v
v


 
2 1
sin sin
   
 (Snell’s law)
3. (ABCD)
Q = n Cv T for isochoric
Q = n CP T for isobaric
1 1
;
V V
V cm P CM
T T
slope
Q nC Q nC
 
 
  
 
Slope(1) =
2
7R
Slope(2) =
2
5R
Slope(3) =
2
3R
4. (ABC)
5. (ABC)
Stress in 2 2
4
3
3
A A
mg
mg
mg
A
r r
 

 
Stress in 2 2
3
3
B B
mg
mg
B
r r
 
 
If A B
r r
 , (stress)A > (stress)B and A will break.
If 2
A B
r r
 , (stress)A = (stress)B

18
Thus, either A or B may break.
6. (BC)
cos
N F 

= 3
2
GM R
x m
R x
 
= 2
constant
2
GMm
R

Between ABC, it executes SHM
At
2
R
x  , acceleration = 0
7. (AD)
2
ˆ
5 /
a i m s

r
10
s
f ma N
 
N = mg = 20 N
Net contact fore = 2 2
10 5
s
N f N
 
8. (B)
1 1 2 2
m v m v

2 1 1 2
/
v m v m
  …………… (1)
2 2 2
2 1 1 2 2
1 1 1
2 2 2 2
a
kx m g m v m v

   ……………… (2) from work energy theorem
Substituting (1) and (2) and solving
 
 
2
2 2
1
1 1 2
4
10 /
4
m ka m ga
v cm s
m m m


 

9. (B)
Mechanical energy loss happens due to viscous/frictional forces without which the liquid will oscillate.
This is released as heat.
10. (C)
Frequencies are in odd ratio hence it is a closed organ pipe
 
2 1
4
v
freq n
L
 
(frequency) =
2
100
4
v
Hz
L

 L = 1.7 m

19
11. (B)
Let block be displaced through x in downward direction and elongation in spring = x1 then
1
cos
x x
  ……………. (1)
2
2 cos
F k x


Hence, 2
2 2 sec
2 cos 2
m m
T
k k
  

 
12. (C)
2
P v


30 /
P
v m s

 
13. (A)
1. (1)
1 1 2 2
2 ,
a R a R
 
 
A B
a a
 (Along the plank)
1 1 2 2
cos 2 cos
a R a R
   
  
Thus a1 = a2
2. (4)
min
3
5 . 5
5
eff
g
v g
   
l l
3gl
3 18
3
5 5
mg mg
T m g T
    
l
Again
4 3 21
5 5 4
mg mg
T
 
    

20
3. (0)
Since force which try to move the system is less than max friction hence system does not move.
Hence, T = 0, f = 0
4. (4)
For solid sphere under pure rolling motion
2
max
5 1
7 2
translational total
KE KE KX
 
 max 4
x cm

5. (8)
Height of water column =
20
4
5
cm

mass of water = 20 gm
by work energy theorem
2
8 10
ext piston water water
W U U KE Joule

     

21
CHEMISTRY:
1. (ABC)
( )
1
P V b RT
PV Pb RT
PV Pb
Z
RT RT
 
 
  
2. (AC)
0 95 25 55 15
o
reaction
H kcal
      
1
30 50 45 40 5
o
reaction
S calK
      
   
3
15 10 400 5
o o o
G H T S ve
            
3. (BCD)
2 2 2 2 2 2
CH CH C CH CH CH CH C CH CH CH CH C CH CH
    
     
    
CH2
or
  or
 
or
  or
 
1 2 3 4 5
6 2
CH 2
CH
1 2
C C
 bonds has double bonds in three of five canonical forms, while 3 6
C C
 has double bonds in
only one canonical form.
4. (BC)
5. (BC)
6. (ABD)
For
COOH COOH
COOH COOK
| & |
When behave as reducing agent
(COOH)2
 2CO2
n-factor = 2, eq wt. = M/2
COOH
COOK
| 
 2CO2 n-factor = 2, eq wt. = M/2
for neutralization meq of one reactant = meq of other reactant can be neutralized
7. (AB)
Acetic acid and O nitrophenol
 are weaker acids than benzoic acid.
8. (D)
9. (A)
The first reaction with diborane is both stereospecific and regiospecific and the second reaction is
stereospecific in that the hydroxyl group replaces the boron from the same side.

22
10. (B)
pH of buffer is given by :
pH = – log Ka
+ log
 
 
Salt
Acid
Case I : 4 = – log 1.0 × 10-5
+ log
 
 
Salt
0.5
or [Salt] = 0.1 × 0.5 = 0.05 M
Case II : 6 = - log 1.0 × 10-5
+ log
 
 
Salt
0.5
log
 
 
Salt
0.5
= 1
[Salt] = 10 × 0.5 = 5 M
Now the two buffer [(I. NaA = 0.05M and HA = 0.5 M) and (II. NaA = 5M and HA = 0.5M)] are
mixed in equal proportion.
Thus, new conc. of NaA is mixed buffer =
0.05 V 5 V 5.05
2V 2
  

New conc. of HA in mixed buffer =
0.5 V 0.5 V
0.5M
2V
  

Thus, pH = – log (1.0 × 10-5
) + log
 
 
5.05 / 2
0.5
pH = 5 + 0.7033 = 5.7033.
11. (D)
4 2
.2
CaSO H O
4 2
1
.
2
CaSO H O
Plaster of Paris
0
120 C
0
200 C
4
CaSO
Dead burnt plaster shortly
heated
2 2
1
2
CaO SO O
 
Calcium Oxide (lime)
12. (C)
   
180
2
3 6
2 C
s v
AlCl Al Cl



(A)
   
2
6
2 6 3
2 6
( ) ( )
H O
s
Al Cl Al OH HCl
A Fumes


  
 
2 6 4 4
3
6 2 6
( )
Al Cl NH OH Al OH NH Cl
A white ppt
   
  2 2
3
Excess
NaOH
Al OH NaAlO H O

 
(Soluble) (C)

23
13. (B)
Equilibria shown for bulb is a mixture of two equilibrium:
     
2
'
decomposition
A s A g H S g

ˆ ˆ ˆ ˆ ˆ ˆ †
‡ ˆ ˆ ˆ ˆ ˆ ˆ
Kp
= 42
= 16
   
sublimation
A s A g
ˆ ˆ ˆ ˆ ˆ †
‡ ˆ ˆ ˆ ˆ ˆ
Kp
= 2
Equilibrium will shift forward to restore the reduced vapour pressure (originally 2) for A. Hence only
sublimation will happen
1. (6)
Number of optically active compounds = 4
Number of meso compounds = 2
Total stereomers = 6
2. (7.35)
S system = nC(v, m)
2
1
3 800
log 2 log
2 100
e e
T
R
T
 
S system = 3R loge 8 = 12.6 cal/k
S surrounding =
3
2 700
2 2.625
800
R
R
 
 
 
  
 
 
 
 – 5.25 cal/k
S total = 12.6 + (–5.25) = 7.35 cal/k
3. (0.8)
4 4 2 4
2 10 8
KMnO FeSO H SO
    
2 4 4 2 4 2
3
2 5 8
K SO MnSO Fe SO H O
  
10 moles of FeSO4 requires 2 moles of KMnO4.
1 mole of FeSO4 require =1/5 moles of KMnO4.
4 2 4 2 4
10 24
KMnO FeC O H SO
    
2 4 4 2 4 2 2
3
3 6 5 20 24
K SO MnSO Fe SO CO H O
    
10 mole of 2 4
FeC O require= 6 mole of KMnO4
1 mole of 2 4
FeC O requires =
6 3
10 5

Total moles of 4
KMnO require =
1 3 4
5 5 5
 
4. (10)
       
2 3 4
0 0 0 0
10 min 2 20 0
0 2 20 0
A g B g C l D s inert gas
t P Pi mm of Hg
t P x x Pi mm of Hg
t P Pi mm of Hg
   

 
 
P + Pi = 210 mm of Hg
P – x + 2x + 20 + Pi = 330 mm of Hg
2P + 20 + Pi = 430 mm of Hg
on solving half-life for A is 10 min
5. (14)
x = 5 fractions, all inactive (achiral (or) racemic)
y = 7 fractions, five active + two inactive (both achiral)
z = 2 fractions, both inactive (one racemic + 1 meso)

24
total 14
Mathematics:
1. (BC)
4
3
b
b a
a
  
B A
  
Using Napier’s analogy
tan
2
tan
2
B A
b a
B A
b a

 
 
  


  
 
 
 cot 7
2
C

  
3
cos cos
4
C A B
  
 C = A – B 
2
A

 (rejected)
C = B – A 
2
B


Let b = 4k, a = 3k & ac = 3
 4
1
7
k 
 circumradius = 4
2
2 7
b

2. (BC)
Terms: 21, 21r, 21r2
, ……………
21r  N &
21
1
S N
r
 

 21 – 21r = 1, 3, 9, 7  
2 2 2
21 3 7
 
Q
 21r = 20  5I
21r = 20 + 18 + 12 + 14 = 64
 
2
min
48
21
7
r  for
4
7
r 
3. (ABD)
2
2 1 3
z z z

3
2
1 2
arg arg
z
z
z z

   
  
   
   

25
3 2
3 2
2 3
2 3
2
arg arg
2 2
1
2
z z
z z
PNB
z z
z z


 
 
 

   
 
  
 
   

 
2
1
1
arg 2
1
z
z

 


 

 
, 3
2
arg
z
z

 

 
 
& 3
1
arg 4
z
z

 

 
 
Using Ptolemy’s theorem for option D
4. (AC)
1
PQ QR
m m
  
2
2 2 3 2 1 1 3 4
t t t t t t t
      ……………… (1)
1 3 4
t t
    
1 2 3 0
t t t
  
Q ……………… (2)
Normal at ‘t’ is y – 2t = –t (x – t2
)
Putting  
: 5 , 4
S  
   
3
2 5 4 0
t t  
    Roots t1, t2, t3
 1 2 2 3 3 1 2 5
t t t t t t 
    ……………… (3)
1 2 3 4
t t t 
  ……………… (4)
 2
t 
 (from (2)), 2
2 2 5 4
t 
    (from (1) & (3))
  = 2, 3
S : (10, –8) or S : (15, –12)
5. (ABD)
Let  
2
sin 0, 1
t x
 
 f(t) = 2 2
1 0
t pt p
   
2 4
5
p 
2
5 4
0 1
2
p p
  
   
0
p 
Q
  
2
2 2
1 1 2
5 5
p p p p
 
        
 
 
    
1 1, 2
p  

26
6. (BC)
 
2 1 4
3
1 4
1
1
z z z
z
z z

 

 1 4 1 4 1
z z z z
  
      
1 4 1 4 1 4 1 4
1 1
z z z z z z z z
    

2 2 2 2
1 4 1 4
1 0
z z z z
   
   
2 2
1 4
1 1 0
z z
  
 4 1
z   
1 1
z 
Q
7. (BC)
   
2
2 1 5
f x x a x a
    
At least one root of f(x) = 0 lies in (1, 3)
 exactly one root lies in (1, 3) & none of ,  = 1 or 3
 f(1) f(3) < 0 
4 8
3 7
a
  
Or both roots lie in (1, 3)  f(1) > 0  f(3) > 0  D  0 
 
2 1
1 3
2
a

  
8
1
7
a
   
Or one root = 1 & other in (1, 3)  not possible
Or one root = 3 & other in (1, 3) 
8
7
a  
4 8 8 8
, , 1
3 7 7 7
a a
   
        
   
   

4
, 1
3
a
 
  
 
 
8. (B)
For the ellipse
2 2
2 2
1
x y
a b
 
     
: cos , sin , , 0 ; , 0
P a b S ae S ae
    
Let incentre of PSS = I(h, k)
PS = a(1 + e cos ); SS = 2ae
Applying formula of incentre
We get h = ae cos ,
sin
1
be
k
e



 locus of incentre is ellipse
 
2 2
2
2 2
2
2
1
1
x y
e
a e
b
e
 

Whose eccentricity is
2
1
e
e

=
2
3
9. (D)
Case I: A selects 9 and B selects any four numbers  8 8
3 4
C C
 ways = n1
Case II: A does not select 9  n2 ways
8 8 8
2 4 4 4
2n C C C
   
 
8 8
4 4
2
1
2
C C
n
 


27
Total number of ways = n1 + n2 =  
8
8 8
4
3 4
2 1
2
C
C C
  
10. (C)
For the ellipse
2 2
2 2
1
x y
a b
 
p = 4x + 4, q = 4x + 2, r = 4x, s = 6x + 3
radius of PAC =
1
2
PI =
1
cos
2 2
P
r ec

where r = inradius of PQR
1
cos 4 5
2 2
P
r ec
 
  
1
tan cos 4 5
2 2 2
p p
s p ec
 
  
 
8 5
qr
s p
s s p
 

(using half angle formula)
 x = 8  sides are of length 32, 34 & 36 perimeter = 102
11. (C)
Given    
2 1, 1
f x x
    where   2
f x ax bx c
  
2
2 4
cx bx a
 
=      
 
   
 
2 1 1
0 1 1 4 0
2
f f
f x f f x f
 
 
    
 
 
=         
2
0 4 1 2 1 2
f x f x f x
     
     
2
0 4 1 2 1 2
f x f x f x
      
2
2 4 2 2 2 2
x x x
     
=  
2
16 2 2, 2
x x
     16
Equality holds if x = 0; f(0) = ± 2; f(1) = 2
m  
1 2
f   m
 for x = 0; a = –4, b = 0, c = 2
or x = 0; a = 4, b = 0; c = –2
12. (B)
Number of ways of arranging 9 boys & 5 girls such that exactly 4 girls are together
5
1 10 4 9
C   
Number of ways out of these in which (B1, B2) are together =  
5
1 9 2 4 8
C    
So required number of ways =  
5
1 9 4 90 16 370 9 4
C  

28

29
13. (B)
Slope of tangent to
2 2
2 2
1
y x
a b
 
at (b tan , a sec ) is
1
sin
2
a
b
  

5
sin
8
   (a = 4, b = 5)

5
tan
39
   ;
8
sec
39
  m
 (m, n) =  
5tan 1, 4sec 2
 
 
=
25 32
1, 2
39 39
 
  
 
 
m (centre of hyperbola is (–1, 2))

7
1
39
m n
  
Numerical Based
1. (33)
Given z50
= 1
49
17 34
1
1
1 k k
k
N
z z


 
 =
  
17
49
17 34 17
1
1
1 1
k
k k k
k
z
z z z


  

=
17
49
51
1
1
1
k
k
k
z
z



 =
17
49
1
1
1
k
k
k
z
z



  
50
1
k
z
 
=
49
2 16
1
1 ......
k k k
k
z z z

   

= 49 – 16 = 33
 
 
49
49
1
1
1
1
k
k
z z
z
z
 



 

 
   
 

 
 

2. (1)
Case I: Let 1
tan x 

  ,
2 2
 

 
 
 
 
 x = tan 
1
tan tan 5
4

 
  
 
 
 
 
 
 
Where
3
4 4 4
  

   
Let ,
4 4 2
  

 
  
 
 
 5
4

 
  
16

  (possible)
Case II: Let
3
,
4 2 4
  

 
  
 
 5
4

  
   
3
16

   (not possible)

30
So, there is only one value of x i.e. tan
16

3. (369)
17 8 12
2 3 5
N 
Perfect squares = n1
   
2 4 6 16 2 4 6 8 2 4 12
1, 2 , 2 , 2 , .........., 2 3 , 3 , 3 , 3 5 , 5 , .........., 5
1 1 8 4 6 8 4 4 6 8 6 8 4 6 315
n              
Perfect cubes = n2
   
3 6 9 12 15 3 6 3 6 9 12
1, 2 , 2 , 2 , 2 , 2 3 , 3 5 , 5 , 5 , 5
2 1 5 2 4 5 2 5 4 2 4 5 2 4 90
n              
Both perfect squares & perfect cubes = n3
    
6 12 6 6 12
3 1, 2 , 2 , 3 5 , 5 1 2 1 2 2 1 1 2 2 2 2 1 2 18
n               
Required no. = 1 2 3
2
n n n
 
= 315 + 90 – 36 = 369
4. (11)
   
, 2, 5
M h k
 
2 2
1
16 9
x y
E    2 2
16, 9
a b
  
Let P, Q, R, S   
,
i i
x y (i = 1, 2, 3, 4)
Slope of PM = slope of normal at P
2
2
i i
i i
y k a y
x h b x

 

2 2 2
2 2 2
1,
x y dy b x
a b dx a y
 
   
 
 
2 2 2 2
i i i i i i
b x y b kx a x y a hy
   
 Points P, Q, R, S lie on the rectangular  
2 2 2 2
0
a b xy b kx a hy
   
Centre 
2 2
2 2 2 2
,
a h b k
a b a b
 

 
 
 
  
16 2 9 5 32 45
, , ,
7 7 7 7
 
  
   
  
   
   
11
 
  
5. (25)
   
14
3 17
0
1 15
k
k
k
k C

 


31
=      
17
3 17
15 16 17
0
1 15 25
k
k
k
k C T T T

     


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