Rankers Academy JEE

FIITJEE – RBT-5 for (JEE-Advanced)

PHYSICS, CHEMISTRY & MATHEMATICS

QP CODE: PAPER - 1
Time Allotted: 3 Hours Maximum Marks: 204

▪ 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.
BATCHES – Two Yr CRP224(All)

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. All the section can be filled in PART-A & B of OMR.
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 Blue/Black Ball Point Pen 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 (01-04) – Contains Four (04) 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–A (05–12) contains Eight (8) Multiple Choice Questions which have One or More Than One
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: −2 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 −2
marks, as a wrong option is also darkened.

(iii) PART – B (1 – 3): This section contains Three (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER. Each question carries +3 marks for correct answer and -1 marks for wrong answer.

(iv) Part-B (4 – 8) contains Five (05) Numerical based questions, the answer of which maybe positive or
negative numbers or decimals TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) and each
question carries +3 marks for correct answer. There is no negative marking.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________

Enrolment Number :_______________________________________________

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 Rankers Academy JEE
IT−2024−RBT-5-Paper-1 (PCM) JEEA

S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – 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. A paper ball (assumed to be carried easily by wind) is to be thrown in a trash can located
6 3
m north of the point of projection. Wind is blowing towards east with speed 2 m/s. The
5
paper ball is thrown at an angle 37 with the horizontal and it falls on the trash can. What is
the speed with which it was thrown?
(A) 4 m/s (B) 5 m/s (C) 6 m/s (D) 7 m/s

2. A very long uniformly charged rod falls with a constant velocity +

+
V through the center of a circular loop making an angle 37° +
+
with axis of loop. Then the magnitude of induced emf in loop is +
+
+
(charge per unit length of rod = ) +

0 0 0
(A) V 2 (B) V 2 (C) V (D) zero
2 2 2

3. In the given figure, the angle of inclination of the inclined plane

V0
is 300. The horizontal velocity V0 so that the particle hits the
inclined plane perpendicularly is H
900
2gH 2gH
(A) v 0 = (B) V0 =
3 7
gH gH 900
(C) V0 = (D) V0 = 300
5 7

4. For a circuit, current variation with time is shown in figure, I0

corresponding rms current is
(A) I0 (B) l0 / 2 0
0 T 2T
t

(C) l0 / 3 (D) I0 / 2
-I0

Space For Rough Work

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PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).

5. A transverse wave is travelling along a stretched string y

from right to left. The figure shown represents the shape

wave motion
of the string (snap shot) at a given instant. At this instant: C
(A) the particles at A, B and H have upward velocity. A
B D
x
(B) the particles at D, E and F have downward velocity. E
H
F
(C) the particles at C, E and G have zero velocity. G
(D) the particles at A and E have maximum velocity.

6. An infinitely long cylindrical conductor of radius R contains current of d

uniform density J along axis of the cylinder. Two infinitely long
cylindrical holes of radius r are drilled symmetrically throughout the
length of the cylinder. The axes of the holes are parallel to the axis of
cylinder and at distance d from it in the same plane.
 J r2 
(A) Magnetic field on the axis of one of the hole will be 0  d − 
2  2d 
0 J  r2 
(B) Magnetic field on the axis of one of the hole will be d − 
2  d
0 J r 2
(C) Magnetic field on the axis of cylinder will be
2 d
(D) Magnetic field on the axis of cylinder will be zero

7. An infinite current carrying wire coming outward passes C

through point O and in perpendicular to the plane
containing a current carrying loop ABCD as shown in the B
figure. Choose the correct option(s).
(A) net force on the loop is zero O O’
(B) Net torque on the loop is zero
(C) As seen from O, the loop rotates clockwise A
(D) As seen from O, the loop rotates anticlockwise D
Space For Rough Work

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8. A particle mass m charge q is moving on a circular vo

P
path on the surface of a frictionless table with speed vo r
where a magnetic field Bo exists uniformly over the
whole region. It is attached by a string which passes
through a hole in the table to a spring as shown.
The spring is stretched by xo. If now the magnetic field
is increased slowly to 2Bo,
(A) The extension in the spring will increase Bo
(B) The speed of the particle will increase
(C) The speed of the particle will decrease
(D) The kinetic energy of the particle will decrease

9. Two projectile are thrown at the same time from two different points. The projectile thrown
from the origin has initial velocity 3iˆ + 3jˆ with respect to earth. The projectile has initial
velocity aiˆ + bjˆ with respect to earth thrown from the point (10, 5). ( î is a unit vector along
horizontal, ĵ along vertical). If the projectile collides after two second, then the
1
(A) value of a is -2 (B) value of a is
2
1
(C) value of b is (D) value of b is –2
2

10. A particle of charge q mass m enters normally (at point P) in V

a region of magnetic field with speed v. It comes out normally
Q
from Q after time T as shown in figure. The magnetic field B
X
is present only in region of radius R and is constant and X
X

uniform. Initial and final velocities are along radial direction B X

X
and they are perpendicular to each other. For this to happen, V X X X
X
which of the following expression(s) is/are correct? P X X X
2mv m m mv X
(A) B  (B) T  (C) T = (D) B  R
qR Bq 2Bq 2qR X
X

11. In the given figure, the block is attached with a system of three
A
ideal springs A, B, C. The block is displaced by a small
distance x from its equilibrium position vertically downwards k

and released. T represents the time period of small vertical

oscillations of the block. Then (pulleys are ideal)
11m
(A) T = 2 B k
2k
(B) the deformation of the spring A is (2/11) times the
displacement of the block.
(C) the deformation of the spring C is (1/11) times the C
displacement of the block. 2k
m
(D) the deformation of the spring B is (4/11) times the
displacement of the block.

Space For Rough Work

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IT−2024−RBT-5-Paper-1 (PCM) JEEA

12. The distance between a screen and an object is 120 cm. A convex lens is placed closed to
the object and is moved along the line joining object and screen, towards the screen. Two
sharp images of the object are found on the screen. The ratio of magnification of two real
images is 1 : 9. Then,
(A) focal length of the lens is 22.5 cm.
(B) smaller image is brighter than the larger one.
(C) larger image is brighter than the smaller one.
(D) brightness of both the images is same.

PART – B (Maximum Marks: 9)

This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.

1. Two blocks A and B, each of mass m are connected by means

of a pulley-spring system on a smooth inclined plane of B K
m
inclination  as shown in the figure. All the pulleys and spring
are ideal. Now, B is slightly displaced from its equilibrium 
position. It starts to oscillate. Time period of oscillation of B is
n, then find n. (Take m = 4 kg, K = 5 N/m,  = 3.14) A
m

2. A small particle slides from height H = 45 cm as shown and B

then loops inside the vertical loop of radius R from where a A
section of angle  = 60° has been removed. If R = (1/N) 60°
H
meter, such that after losing contact at A and flying through R
the air, the particle will reach at the point B. Find N. Neglect
friction everywhere.

3. A thin circular disc of radius R is made to rotate with a

constant angular speed  within a oil filled (coeff. of
viscosity  ) cylindrical box as shown in the figure. The
clearance between the disc & the horizontal planes of
the cylindrical box is very small & is equal to h.
Considering that the vertical side of the cylindrical box
is almost in contact with the disc, the power to be
supplied to the system to maintain the constant
xR y
angular speed is . Then x + y is
h
Space For Rough Work

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IT−2024−RBT-5-Paper-1 (PCM) JEEA

PART – B (Maximum Marks: 15)

(Numerical Type)
This section contains Five (05) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

4. A ring of mass 3.15 kg is rolling without slipping with linear velocity 1 m/sec on a smooth
horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic
energy of the system (in J).

x3 9x2
5. Potential energy of a particle moving along x-axis is given by U = − + 20x . Find out
3 2
position of stable equilibrium state.

6. As shown in figure light string PQR is stretched by P

force F = (3 – 10kt) N, where k is a constant and t is Q
time in second. At time t = 0, a pulse is generated at
the end P of the string. For the value of k (in N/s) if the R
value of force becomes zero as the pulse reaches L = 1.6 m
point Q. Given mass per unit length of string is
0.03 kg/m. F = [3-k(10)t]N

7. In a moving coil galvanometer, torque on the coil can be expressed as   i, where i is

current through the wire. The rectangular coil of the galvanometer having number of turns N,
area A and moment of inertia  is placed in magnetic field B. The torsion constant of spring in
the galvanometer is C and charge Q is passed through the coil almost instantaneously.
KNABQ
If the maximum angle through which coil deflected is . (Ignore the damping in
3C
mechanical oscillations). Then the value of K is

8. 2 kg of ice at –22.50C is mixed with 2.5 kg of water at 250C in an insulating container. If the
specific heat of ice and water are 0.5 cal/gm°C and 1 cal/gm°C respectively. Latent heat of
fusion of ice = 80 cal/gm. Find the amount of water present in the container (In kg).

Space For Rough Work

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IT−2024−RBT-5-Paper-1 (PCM) JEEA

S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – 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. Choose correct statement from the following

(A) Bond angle: OCl2 > BCl3 > NCl3
(B) Strength of backbonding: BCl3 > OCl2 > NCl3
(C) Bond order: NCl3 > OCl2 > BCl3
(D) Dipole moment: NCl3 > BCl3 > OCl2

2. R − C  C − R ⎯⎯⎯→
LiAlH4
RCH = CHR
In the above reaction, R and R should be
(A) R = CHO and R = CH3 (B) R = CH3 and R = CH2OH
(C) R = C2H5 and R = CO (D) R = CH3 and R = COOH

3. The S – O bond order of SOF2 is greater than that of SOCl2. This is due to
(A) polarity of O – F bond is greater than that of O – Cl bond
(B) the O → S back bond in SOF2 is stronger than that in SOCl2
(C) the S → O back bond in SOF2 is stronger than that in SOCl2
(D) the FSF is greater than ClSCl

4. Which alcohol can’t be formed by acidic hydration of any alkene?

OH
(A) (B) OH

OH
(C) (D) OH

Space For Rough Work

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PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).
5.

6. Compound A(C5H7OCl) reacts rapidly with ethanol and catalytic amount of acid to form a
pleasant smelling substance B(C7H12O2). A also reacts with H2O to form C with neutralization
equivalent of 100. A, B and C all react with Br2 water. Acid C, which can also be obtained by
acidic hydrolysis of B is oxidized to new acid D(C4H6O3) and CO2. D on heating with soda
lime gives acetone. Identify A to D.
(A) A = CH3 - C - CH 2 - C- Cl (B) B = CH3 - C - CH 2 - C - OEt

CH2 O CH2 O
(C) C = CH 3 - C - CH 2 - C - COOH (D) D = CH 3 - C - CH 2 - COOH

CH2 O O

7. Choose the correct statement

(A) dyz orbital lies in the xz plane
(B) pz orbital lies along the x axis
(C) lobes of dx 2 − y 2 orbital are at 90o with the z axis
(D) lobes of dxy orbital are at 90o with the z axis
Space For Rough Work

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8. The spectra of which complex(s) is/are spin forbidden as well as Laporte forbidden?
(A) [Mn(H2O)6]2+ (B) [Fe(H2O)6]3+
3-
(C) [Co(CN)6] (D) [Cr(NH3)6]3+

9. A hydrocarbon X (C10H16) upon catalytic hydrogenation gives 4-methyl-1-isopropyl

cyclohexane. Also (X) upon ozonolysis followed by hydrolysis in the presence of Zn gives
CH2O and

O O

H3C CH3

CHO

The correct statement(s) concerning (X) is/are

(A) CH3 (B) (X) has two chiral carbon.
Structure of (X)His
3C
CH2
(C) (X) has one chiral carbon. (D) With excess of HCl, (X) gives
racemic dichloride.

10. Which of the following reactant(s) produce(s) racemic mixtures when treated with HBr?
(A) Cis-2-butene (B) Trans-2-butene
(C) 1-butene (D) Isobutene

11. Which of the following increases by increasing the temperature of a chemical reaction?
(A) Rate constant (B) Half-time
(C) Rate of reaction (D) Temperature coefficient

12. Which reaction produces cyclic products?

COONa
CH2
⎯⎯⎯⎯⎯⎯⎯
Kolbe ' s electrolysis
→
(A) CH3COCH3 ⎯⎯⎯⎯⎯
Conc.H2SO4
Heat
→ (B) CH2
CH2
COONa
OH
o

(C) C6H5COONa ⎯⎯⎯⎯

Soda lim e
Heat
→ (D) CH3 − C − CH3 ⎯⎯⎯⎯→
Cu/300 C

CH3

Space For Rough Work

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PART – B (Maximum Marks: 9)

This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.

1. OCH 3
OCH 3
HI
⎯⎯⎯ → Products

OCH 3
OCH 3
If x moles of HI can be completely consumed by one mole of the above compound, what is
the value of x?

2. FeCl3 + K4[Fe(CN)6] → Fe4[Fe(CN)6]3

If ‘x’ moles of KSCN will be completely consumed by one mole of product in above reaction,
to produce blood red colouration, what is the value of x/3?

3. If the pH of 0.1 M aqueous solution of NH4NO3 is x, what is the value of x? [Kb of NH4OH =
10–5]

PART – B (Maximum Marks: 15)

(Numerical Type)
This section contains Five (05) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

4. 2.5 mole of KCl is added to one Kg of water at 1 atm. At what temperature in kelvin unit will
the water boil if the degree of dissociation of KCl is 0.8? [Kb of H2O = 0.52 K kg mol–1]

5. 14 g of a metal oxide of formula MO is added to a container containing 500 mL of 1.25 M

HCl solution. After complete reaction, the reaction mixture is treated with 250 mL of 0.5 M
NaOH to neutralize the excess acid. If the atomic mass of M is expressed as (2x + 15). What
is x?

6. The successive ionization energies of a s-block element are

8.6, 12.9, 1630.2, 2406.2 eV, etc
If atomic mass of the elements is 39.8 g mol–1, what will be the molar mass of it’s normal
oxide?
Space For Rough Work

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7. The molecular mass of a polyhydric aliphatic alcohol, increases by 252 unit if it reacts with
CH3COCl/(CH3CO)2O. if the number of OH groups present in the alcohol is x, what is the
x
value of ?
4

8. A 2 ( g) ⎯⎯→ A ( g) + A ( g )
The heat of reaction of above decomposition of a diatomic molecule A2(g), into monoatomic
gases A(g) is 40.5 kJ mol-1 at 320 K. What will be the enthalpy change of the reaction at 330
K in kJ mol–1 unit?
5R
[CP of monoatomic gases = ]
2
7R
[CP of diatomic gases = ]
2
[R = 8.3 J K-1 mol-1]
Space For Rough Work

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IT−2024−RBT-5-Paper-1 (PCM) JEEA

S
SEEC
CTTIIO
ONN –– IIiiii :: M
MAAT
THHE
EMMA
ATTIIC
CSS
PART – 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. Let a, b, p, q and suppose that f ( x ) = x2 + ax + b = 0 and g ( x ) = x3 + px + q = 0 have a

common irrational root, then
(A) f ( x ) divides g ( x ) (B) g ( x )  xf ( x )
(C) g ( x )  ( x − b − q) f ( x ) (D) none of these

( ) =  x  + f, then x (1 − f ) is equal to
2n
2. If x = 7 + 4 3
(A) 1 (B) 2
(C) 3 (D) 4

3. Consider the points P = ( − sin ( −  ) , − cos  ) ,Q = ( cos ( −  ) ,sin  ) and


R = ( cos (  −  +  ) , sin ( −  ) ) , where 0  ,   then
4
(A) P lies on the line segment RQ (B) Q lies on the line segment PR
(C) R lies on the line segment QP (D) P, Q, R are non – collinear

4. For a  b  c  0, the distance between (1, 1) and the point of intersection of the lines
ax + by + c = 0 and bx + ay + c = 0 is less than 2 2 . Then
(A) a + b − c  0 (B) a − b + c  0
(C) a − b + c  0 (D) a + b − c  0

PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).

5. The number of triangles that can be formed with the sides of lengths a, b and c where a, b, c
are integers such that a  b  c is
1 1
(A) ( c + 1) when c is odd (B) c ( c + 1) when c is odd
2

4 2
1 1
(C) c ( c + 2 ) when c is even (D) c 2 when c is even
4 4
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1+ x 1 + 2x 1 + 3x
6. If x  R, and S = 1 − C1 + C2 − C3 + ... + .. upto (n + 1) terms then S
1 + nx (1 + nx ) (1 + nx )
2 3

(A) equals x 2 (B) equals 1

(C) equals 0 (D) is independent of x

7. A and B are two points on the x – axis and y – axis respectively. Two circles are drawn
passing through the origin and having centre at A and B.
(A) Equation of the common chord is ax − by = 0
 ab2 a 2b 
(B) mid – point of the common chord is  2 . 2 
a +b a +b 
2 2

(C) AB bisect the common chord.

(D) AB is perpendicular to the common chord.

b cos x b + sin x
8. The equation = posses a solution if
(
2 cos 2x − 1 cos x − 3 sin2 x tan x
2
)
1 1 1
(A) b  (B) 0  b  (C) b  0 (D) b =
2 2 2

9. Normals are drawn from the point P (15, 12) to the parabola y2 = 4x . If the feet of the
normal form a ABC, then which of the following is/are true.
 26 
(A) Centroid of ABC is  , 0 
 3 
 22 
(B) Centroid of ABC is  , 0 
 3 
(C) Area of triangle formed by tangents to the parabola at A, B and C is 35 sq. units
(D) Area of triangle formed by tangents to the parabola at A, B and C is 140 sq. units

10. Let z 1 lies on z = 1 and z 2 lies on z = 2 , then which of the following is/are true?
(A) maximum value of z1 + z2 is 3 (B) minimum value of 2z1 − z2 is 0
(C) maximum value of 2z1 + z2 is 4 (D) minimum value of 2z1 − 3z2 is 5

11. If pair of tangents are drawn from centre C of the circle x2 + y2 − 5x = 0 to the nearer branch
x2 y2
of the hyperbola − = 1 to touch it at point A and B, then
25 16
−64
(A) product of the slopes of lines CA and CB is
75
−32
(B) product of the slopes of lines CA and CB is
25
(C) circumcentre of ABC lies inside the ABC .
(D) circumcentre of ABC lies outside the ABC .
Space For Rough Work

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12. If a + b + c = 4 and a2 + b2 + c 2 + 3 (ab + bc + ac ) = 21 where a,b,c  , then which of the

following is/are true
(A) minimum value of ( a + b)(b + c )(a + c ) is 18
(B) minimum value of ( a + b)(b + c )(a + c ) is 24
490
(C) maximum value of ( a + b)(b + c )(a + c ) is
27
471
(D) maximum value of ( a + b)(b + c )(a + c ) is
27

PART – B (Maximum Marks: 9)

This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE
INTEGER.


1. Consider the number N = 10! . If the number of positive divisors of N is  , then equals
30

2. Let L1,L 2 and L 3 be the length of tangents drawn from a point P to the circles x2 + y2 = 4 ,
x2 + y2 − 4x = 0 and x2 + y2 − 4y = 0 respectively. If L14 = L22L32 + 16, then locus of P are the
curves C1 (a straight line) and C2 (a circle). Let the circumcentre of the triangle formed by C1
and other two lines which are inclined at an angle of 45o with C1 and is tangent to C2 is
( ,  ) , then (  + 2) equals

3.
n
Let Sn =  r .2 2
r
(1+( −1) ).3 2r (1−( −1)r ) . If
r

S20 =  .321 +  .222 +

391  32 − 9 
, then the value of   is
r =1 288  4 
(where [.] denotes the greatest integer function)

PART – B (Maximum Marks: 15)

(Numerical Type)
This section contains Five (05) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

4. If 16cos x + 12sin x  = 4k + 10, then the maximum value of k is _______________

Space For Rough Work

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1
5. The length of the semi – major axis of an ellipse is 8 and the eccentricity is . If  denotes
2
the area of the rectangle formed by joining the vertices of the latera recta of the ellipse then

the value of is equal to
120

6. The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from
each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and
C be a circle touching C1 and C2 externally. If a common tangent to C1 and C passing
through P is also a common tangents to C2 and C1 , then the one fifth of the radius of the
circle C is

1
7. Value of x if x satisfies the equation log1− x ( 3 ) − log1− x ( 2 ) = is
2

5
The sum of all the integral roots of (log5 x ) + log5x   = 1 is ______________
2
8.
x
Space For Rough Work

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QP Code:
Answers
S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – A

1. B 2. D 3. A 4. C
5. ABD 6. AD 7. AC 8. ACD
9. AC 10. ABC 11. ABCD 12. AB

PART – B
1. 2 2. 5 3. 6 4. 5.25
5. 5 6. 1.25 7. 1.73 8. 3

S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – A
1. B 2. B 3. B 4. C
5. ACD 6. ABD 7. CD 8. AB
9. AC 10. ABC 11. ACD 12. ABC
PART – B
1. 7 2. 4 3. 5
4. 375.34 (Range 375.2 to 375.4) 5. 12.5 (Range12.4 to 12.6)
6. 55.8(Range 55.7 to 55.9) 7. 1.5
8. 40.62(Range 40.6 – 40.7)

S
SEEC
CTTIIO
ONN –– IIIIII :: M
MAATTH
HEEM
MAATTIIC
CSS
PART – A
1. A 2. A 3. D 4. A
5. AC 6. CD 7. ABCD 8. ABC
9. AC 10. ABC 11. AC 12. AC
PART – B
1. 9 2. 0 3. 4 4. 2.50
5. 0.8 6. 1.6 7. –1.25 8. 6

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Answers & Solutions

S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – A
1. B
Sol. From range:
2uxuy 6 3
= ux uy = 6 3 …(1)
g 5

Resultant of 2 and ux  4 + u2x

Whose resultant with uy makes an angle 37°.
u2y 9
So, = …(2)
4 + u2x 16

Solving (1) + (2), we will get ux = 2 3, uy = 3 & uz = 2 m / s

So net  u2x + u2y + u2z = 5 m/s.

2. D
Sol. B = 0
dB
 B⊥S ;  =0
dt

3. A
Sol. vy = 2gH V0

vy 2gH H
= = tan 60o = 3
90°

vx vo Vo
60°
Vy V
2gH
v0 = 30°
3

4. C
2Io
Sol. i= t [0 < t < t/2]
T
T/2
2Io
 t dt
2
T 0 Io
irms = = .
T/2 3

5. ABD
Sol. Results can be obtained by looking at shape of string
after a short time (shown dotted). Also, each particle
of string executes SHM about mean position which is
on x-axis.

6. AD
Sol. Use ampere`s law to find magnetic field due to solid cylinder current = µ0Jd/2
Due to cylinder on its own axis is 0
Applying superposition principle we can get the required answers.

7. AC
Sol. Force on branch BC is coming out of the plane of paper and force on AD is going into the
plane of paper and net force on the loop is zero.

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8. ACD
Sol. ( )
Fm = q v  B and apply Newton’s second law of motion.

9. AC
Sol. b 3-a
w.r.t. A
B
a 3-b

3 5
3 5
A
(0, 0) 10

(0, 0) 10

 d
In vertical direction  s = 
 t
5 1
3−b = → b =
2 2
In horizontal direction
10
3−a = → a = −2.
2

10. ABC
Sol. So, its radius is also R.
mv mv
So, R = → B=
qB qR
2M m
T= =
4qB 2qB
So, A, B & C are correct.

11. ABCD
Sol. Let elongation in spring A, B and C be x1, x2 and x3 respectively.
Considering spring forces and constraint relations
x2 = 4x3 …(i)
x2 = 2x1 …(ii)
and x1 + 2x2 + x3 = x …(iii)
2 4  1
 x1 =   x ; x 2 =   x ; x 3 =   x
 
11  
11  11 
 x 
Also, F = 2K  
 11 
11m
 T = 2
2k

12. AB
Sol. For two different positions of lens
u 120 − u
u 120 – 4 1= ; 9=
120 − u u
2
1 u 1 u
= → =
9 (120 − u) 2
3 120 − u
u = 30 ; v = 90
120

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1 1 1 1 1
= − = +
f v u 90 30
P = 22.5

PART – B
1. 2
Sol. Let elongation of spring be x0 in equilibrium. Then,
2T + mgsin  = 2kx0 …(i)
and T = mg …(ii)

2k(x0 + 2x)
Let Block B is displaced by x down the inclination
F.B.D. of B aB
−maB = 2k(x0 + 2x) − 2T − mgsin  …(iii)
2T 
F.B.D. of A mg sin

mg – T  = maA T
Also, a A = 2aB
T = mg − 2maB aA
−maB = 2kx0 + 4kx − 2mg + 4maB − mgsin 
−maB = 4kx + 4maB m
g
4k
aB = − x
5m
5m
 T = 2
4k
T = 6.28 s.

2. 5
Sol. Conservation of mechanical energy
1
mgH = mgh + mv 2
2 V
R 1
0.45 mg = 3mg + mv 2
B

2 2 R/2
60°
A
9 – 30R = v2 …(i) C
60°
R 3
AC =
2
. For reaching at B let particle takes t sec.
v R 3
t = …(ii)
2 2
v 3 1 R
t −  g  t2 = …(iii)
2 2 2
On solving, v2 = 15R by eq. (i)
9
9 – 30R = 15R  R=  100 = 20 cm
45
= 1/5 meter.

3. 6
 r 
Sol. dF =  ( 2rdr )  2
 h
dP = dF. V

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4 
( r ) r dr
2
=
h
P 42 R 3 2R 4
o dP =
h o
r dr =
h
x+y=6

4. 5.25
1 1 1 m(2r)2 2 1 2 2
Sol. K.E. = mv 2 + mv 2 + w + mr w = 5.25.
2 2 2 12 2

5. 5
Sol. At stable equilibrium, U is minimum.
dU d2U
= 0 and >0
dx dx 2
1  x3 ax 2 
=  − + 20x  = 0.
dx  3 2 

 x – 9x + 20 = 0.  (x – 5) (x – 4) = 0.
2

x = 5 and x = 4 are points of equilibrium.

d2U
And U minimum when > 0. i.e. at x = 5.
dx 2
6. 1.25
dx T 3 − 10kt
Sol. For pulse = =
dt  
to
3 − 10 kt
L

  dx =  dt
o o

And 3 – 10 k to = 0

7. 1.73
Sol.  = NIAB = C
NABQ
=
C
So, K = 3
K = 1.73

8. 3
Sol. 2000 × 15 × 0.5 + m × 80 = 2500 × 25 × 1
m = 500 gm
Water = 2.5 kg + 0.5 kg = 3 kg.

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S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – A
1. B
Sol. Back bonding in BCl3 is easier as it is electron deficient compound. Between OCl2 and NCl3
the electronegative difference between atoms in OCl2 is greater than between atoms in NCl3.
OCl2 forms stronger back bond than NCl3.

2. B
Sol. R or R can’t be CHO, CO, COOH as they are reduced by LiAlH4.

3. B
Sol. The O → S back bond in SOF2 is stronger than O → S in SOCl2. Since F is more
electronegative than Cl, the S atom in SOF2 strongly attract the lone pair of oxygen to form
p - d back bond.

4. C
Sol. No alkene can produce the alcohol in (C) because loss of H2O can’t form any alkene.

5. ACD
Sol. This is Hoffmann’s bromamide reaction.

6. ABD
Sol. The alkene part of the carboxylic acid derivatives reacts with Br2/H2O.

7. CD
Sol. dx 2 − y 2 lie along X- and Y- axis, which are perpendicular to Z-axis.

8. AB
Sol. Due to t 32geg2 crystal field electronic configuration of Mn2+ and Fe3+ ions

9. AC
Sol. Cleavage of C = C bond takes place during ozonolysis.

10. ABC
Sol. Racemic mixture is formed from compounds containing chiral carbon atom.

11. ACD
Sol. t½ only decreases by increasing temperature.

12. ABC
Sol. In A, B and C the products are mesitylene, cyclohexane and benzene respectively.

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PART – B
1. 7
Sol. OCH 3 OH
OCH 3 OH
4 HI
⎯⎯⎯⎯ → + 2 CH3I
OCH 3 OH
OCH 3 OH 2 HI

OH OH OH
I I
⎯⎯ HI
−I2
⎯⎯ ⎯ +2H2O
I
OH OH OH

2. 4
Sol. 4Fe3+ + 12SCN− ⎯⎯
→ 4[Fe(SCN)3 ]
One mole Fe3+ consumes three moles of SCN– ion. Four moles of Fe3+ will consume 4  3 =
12 moles Fe(SCN)3.
x=4

3. 5
1 kw 1 1
Sol. pH = p − pkb − logC = 14 − 5 − log10 −1  =  10 = 5
2  2  2

4. 375.34 (Range 375.2 to 375.4)

Sol. Tb = iKbm = 1.8  0.52  2.5 = 2.34
 Tb - Tb0 = 2.34
or Tb – 100oC = 2.34
 Tb = 102.34oC = 375.34 K

5. 12.5 (Range12.4 to 12.6)

Sol. Meq of HCl added = 500  1.25 = 625
Meq of NaOH used = 250  0.5 = 125
Meq of HCl reacted with MO = 625 – 125 = 500
W 14
500 = Meq of MO =  1000 =  1000  E = 28,molar mass = 28  2 = 56
E E
At weight of MO = 56 – 16 = 40
 2x + 15 = 40
 x = 12.5

6. 55.8(Range 55.7 to 55.9)

Sol. The atom contains two unpaired electrons.
 Formula of oxide = MO
Molar mass = 39.8 + 16 = 55.8

7. 1.5
O
Sol. ( CH CO) O
ROH + CH3COCl ⎯⎯⎯⎯⎯
3 2
→ R − O − C − CH3 + HCl
For one OH group (OH → OCOCH3), loss of one H and gain of COCH3 takes place.

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 For one OH group, the mass increases by 42 units.

252
 Number of OH group = =6=x
42
x 6
 = = 1.5
4 4

8. 40.62(Range 40.6 – 40.7)

Sol. H2 = H2 + nCPdT
330
= 40.5 + nCP  dT
320

= 40.5 + 0.1(1.5 R) (330-320)

= 40.5 + 1.5R  10 = 40.5 + 15R
= 40.5 + 15  8.3 = 40.62

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S
SEEC
CTTIIO
ONN –– IIIIII :: M
MAATTH
HEEM
MAATTIIC
CSS
PART – A
1. A
Sol. Let   − be a common root of f ( x ) = 0 and g ( x ) = 0 . Then 2 = −a − b . Substituting
this in 3 + p + q = 0, we get ( a2 − b + p )  + ab + q = 0
As  is irrational and a,b,p,q  , p = b − a2, q = −ab . This gives, g ( x ) = ( x − a) f ( x ) .

2. A
1
Sol. We have 7 − 4 3 =
7+4 3

( )
2m
0  7 − 4 3  1  0  7 − 4 3 1

( )
2n
Let F = 7 − 4 3 .

( ) + (7 − 4 3 )
2n 2n
Then x + F = 7 + 4 3

= 2  2n C0 72n + 2nC2 72n−2 (4 3 ) + C 7 (4 3 ) ( ) 

2 4 2n
2n − 4
2n
+ ... + 2nC2n 4 3
 4 
= 2m, where m is some positive integer.
  x  + f + F = 2m  f + F = 2m −  x 
Since 0  f  1 and 0  F  1, we get 0  f + F  2 . Also, since f + F is an integer, we
must have f + F = 1. Thus, x (1 − f ) = xF = 7 + 4 3 ( ) (7 − 4 3 )
2n 2n

= ( 49 − 48 )
2n
= 12n = 1.

3. D
− sin  − cos  1
Sol. Put  −  −  and consider the determinant  = cos  sin  1
cos (  +  ) sin (  −  ) 1
Using R3→ R3 − cos R2 − sin  R1
− sin  − cos  1
 = cos  sin  1
0 0 1 − cos  − sin 
= (1 − cos  − sin ) cos (  +  )
= (1 − cos  − sin ) cos ( 2 −  )
   
= 1 − 2 sin   +   cos ( 2 −  )
  4 
   
As 0       + 
4 4 4 2
1    − 
  sin   +   1 and 0  ,     2 −  
2  4 4 4 2
 cos ( 2 −  )  0
Thus   0 and the points P, Q, R are non – collinear.

4. A
 −c −c 
Sol. The lines ax + by + c = 0 and bx + ay + c = 0 intersect in  , .
a+b a+b

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 −c −c 
Distance between (1, 1) and  , 
a+b a+b
c a+b+c
= 2 1+ = 2 .
a+b a+b
a+b+c
As 2  2 2, we get a + b + c  2 ( a + b )
a+b
a+b−c 0.

5. AC
Sol. Let c = 2m + 1 where m is a positive integer. Since c  a + b and a  b, we get
1 1
c  2b  b  c = m + . Thus, b can take values from m + 1 to 2m + 1.
2 2
If b = 2m + 1, a can be 1,2,...... ( 2m + 1) . Thus, there are ( 2m + 1) values.
If b = 2m, a can be 2, 3,….2m. In this case a can take ( 2m − 1) values. And so on. When
b = m + 1, a can take just one value viz. (m + 1) . Thus, there are
( 2m + 1) + ( 2m − 1) + .... + 1 = (m + 1)
2
.

6. CD
1
= y, we can write S = C0 − C1 y + C2 y 2 − ... + ( −1) Cn y n 
n
Sol. Putting
1 + nx  
− xy C1 − 2C2 y + 3C2 y 2 − ... + ( −1) nCn y n−1 
n −1

 

= (1 − y ) + xy
n d
dt

(1 − t ) t = y
n

= (1 − y ) − nxy (1 − y )
n n −1

= (1 − y ) 1 − y − nxy 
n −1

= (1 − y ) 1 − (1 + nx ) y  = 0 .
n −1

7. ABCD
x y
Sol. + = 1. Equations of
Let OA = a and OB = b (figure), so that the equation of the line AB is
a b
the circles passing through the origin and having centres at A (a, 0) and B (0, b) are,
respectively.
x2 − 2ax + y2 = 0 (1)
and x2 − 2by + y2 = 0 (2)
so that the equation of the common chord OP is
ax − by = 0 (3)
which is perpendicular to AB
This has one end at the origin O and the other end P is given by solving (2) and (3).
2
b 
 a y  − 2by + y = 0
2

 
2a2b 2ab2
y= 2 x= 2
a +b 2
a + b2

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P
(0, b)
B

O (0, 0) A (a, 0)

 ab2 a 2b 
Therefore, the mid – point of the common chord OP is  2 , 2 
which lies on AB
a +b a +b 
2 2

8. ABC
Sol. The conditions for the existence of a solution are
1 
1. 2cos 2x − 1  0  cos 2x   2x 
2 6

2. tan x  0  x  
2
1 
3. cos2 x − 3 sin2 x  0  tan2 x   x  
3 6
As 2 cos 2x − 1 = 2 ( cos x − sin x ) − 1
2 2

( ) (
= 2 cos2 x − sin2 x − cos2 x + sin2 x )
= cos2 x − 3 sin2 x
So the given equation can be written as b sin x = b + sin x
b
 sin x =
b −1
b
Which is possible if −1  1
b −1
2b − 1 1
  0 and 0
b −1 b −1
1
 2b − 1  0  b 
2
1 −
But for b = ,sin x = −1 which is not possible as x  .
2 2
1
Hence equation has a solution if b  .
2

9. AC
Sol. tx + y = 2t + t3 passes through (13, 12) t3 − 13t − 12 = 0
 t 2 ( t + 1) − t ( t + 1) − 12 ( t + 1) = 0
( t + 1)( t − 4)( t + 3) = 0
A, B, C = ( t 2 ,2t )
A (1, − 2) ; B (16,8) ;C (9, − 6 )
1 −2 1
 26  1
Centroid =  , 0  , Area = 16 8 1
 3  2
9 −6 1
A = 70

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70
Area of triangle formed by tangents = = 35 .
2

10. ABC
Sol. z1 = 1 and z2 = 2
(A) z1 + z2  z1 + z2  3
(B) 2z1 − z2  2 z1 − z2  0
(C) 2z1 + z2  2 z1 + z2  4
(D) 2z1 − 3z2  2 z1 −3 z2  4

11. AC
5 
Sol. centre:  , 0  ; y = mx  25m2 − 16
2 
5 25m2
m =  25m2 − 16  = 25m2 − 16
2 4
75m2 64 64
= 16  m2 − = 0, product of slopes of CA . CB is − circumcentre of ABC lies
4 65 75
inside the ABC

12. AC
Sol. ( a + b ) + (b + c ) + ( a + c ) = 8
Let a + b = ,b + c = ,a + c = 
++  =8
  +  +  = a2 + b2 + c 2 + 3 (ab + bc + ac )
= 21
 x3 − 8x 2 + 21x −  = 0 where  = ( a + b )(b + c )( a + c ) these equations have 3 real roots.
490
27

18

 3
7/3
f ( x ) = xe , f ' ( x ) = − ( x − 1) e ,f " ( x ) = e ( x − 2)
−x −x −x

g ( x ) = xex ,g' ( x ) = ( x + 1) ex ,g" ( x ) = ( x + 2) ex

p ( x ) = x3 − 3x,p' ( x ) = 3x2 − 3,p" ( x ) = 6x
q ( x ) = x4 − 2x2 + 3,q' ( x ) = 4x ( x − 1)( x + 1) ,q" ( x ) = 12x2 − 4

f'(x) f'’(x) 1/e

h (x) = f(x)
+ – + –
, ,
(1, 0)
1 2

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g'(x) g'’(x)

– + + (–1, 0) h (x) = g(x)

, – ,

–1 –2

–1/e

p'(x) p'’(x) 2

+ – + + h (x) = p(x)
, – ,
–1 1 (–1, 0) (1, 0)
0

–2
(0, 3)

q'(x) q'’(x)
(0, 2)
– + – + + – +
, ,
−1 1 –1 0
–1 0 1 1
3 3
(I), (i) (iv), (R)
(II), (i) (iv), (R, S)
(III), (ii) (iv), (P, Q, R)
(IV), (ii) (iii), (P, S)

PART – B
1. 9
Sol. 10! = 28.34.52.7
Number of divisors = 9  5  3  2 = 270 = 

2. 0
(h ) = (h )( )
2
Sol. 2
+ k2 − 4 2
+ k 2 − 4h h2 + k 2 − 4k + 16

(
 ( h + k ) h2 + k 2 − 2h − 2k = 0 circle ) 45o

C1 : x2 + y2 − 2x − 2y = 0 and
C2 : x + y = 0 circumcentre will be mid point of (1, 1)

hypotenuse in right angle triangle ( , ) = ( 0, 0)

x+y=0

3. 4
Sol. Sn = 1.3 + 2.22 + 3.33 + 4.24 + ... + upto n – terms
( ) (
Sn = 1.3 + 3.33 + 5.35 + ... + 2.22 + 4.24 + 6.26 + ... )
S1 = 1.3 + 3.33 + 5.35 + ... + upto 10 – terms, S1 =
15
32
5.321 + 1 ( )
58.222 + 8
S2 = 2.22 + 4.24 + 6.26 + ... + upto 10 – terms, S2 =
9

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 58.222 + 8 
15
S20 = S1 + S2 = 5.321 + 1 + (  )
32  9 
75 21 58 22 391
= .3 + .2 +
32 9 288

4. 2.50
3
Sol. 16cos x + 12sin x = 162 + 122 cos ( x −  ) ,  = tan−1   .
4

5. 0.8
x2 y2  1
Sol. Let the equation of the ellipse be 2
+ 2 = 1 where a2 = 82 and b2 = 82  1 − 
a b  4
 b2 
Vertices of the letra recta are  ae,  
 a
2b2
So  = 2ae  = 4b2 e
a
3 1
= 4  82   = 12  8
4 2

 = 0.8
120

6. 1.6
C
Sol. Let A 1, A 2 and M be the centres of the circles
C1,C2 and C respectively. Let the common
tangent through P to C1 and C touch C1 at
B1, C at B 2 and C2 also at B 2 . M

From right angled triangle A 1B1P if B2

AB 1
A1PB1 = ,sin  = 1 1 =
A1P 3 A2 C2

2 2
 cos  =  PB1 = 2 2 = PB2
3 P
PB2 2 2
From triangle MPB2 , tan  = = A1

MB2 r

B1
12 2
 = = r = 8 . C1
2 2 r

7. –1.25
Sol. 1 − x  0, 1 − x  1  x  1, x 0
3 1 3
Also, log1− x   =  = (1 − x )
1/2

2 2 2
9 9 −5
 = 1− x  x = 1− = .
4 4 4
 −4x = 5
 x = −1.25

8. 6
1
Sol. Clearly x  0 and x 
5
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 5  log5 5 − log5 x
log5x   =
 x  log5 5 + log5 x
Putting log5 x = t, then equation (1) becomes
1− t
t2 + = 1  t 3 + t 2 − 2t = 0
1+ t
 t ( t − 1)( t + 2) = 0  t = 0, 1, − 2
So integral roots of (1) are 1 and 5.

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