29-01-2025

1001CJM290125MRG JM

PART-1 : PHYSICS

SECTION-I

1) Given below are two statements : one is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) : Choke coil is simply a coil having a large inductance but a small resistance. Choke
coils are used with fluorescent mercury-tube fittings. If household electric power is directly
connected to a mercury tube, the tube will be damaged.
Reason (R) : By using the choke coil, the voltage across the tube is reduced by a factor

, where ω is frequency of the supply across resistor R and inductor L. If the choke
coil were not used, the voltage across the resistor would be the same as the applied voltage.
In the light of the above statements, choose the most appropriate answer from the options given
below:

(A) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(B) (A) is false but (R) is true.
(C) Both (A) and (R) are true and (R) is the correct explanation of (A).
(D) (A) is true but (R) is false.

2) Two projectiles are fired with same initial speed from same point on ground at angles of (45° – α)
and (45° + α), respectively, with the horizontal direction. The ratio of their maximum heights
attained is :

(A)

(B)

(C)

(D)

3) An electric dipole of mass m, charge q, and length l is placed in a uniform electric field .
When the dipole is rotated slightly from its equilibrium position and released, the time period of its
oscillations will be :

(A)

(B)
(C)

(D)

4) The pair of physical quantities not having same dimensions is :

(A) Torque and energy

(B) Surface tension and impulse
(C) Angular momentum and Planck's constant
(D) Pressure and Young's modulus

5) Given below are two statements : one is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the
base of the mountain.
Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due
to gravity and vice-versa.
In the light of the above statements, choose the most appropriate answer from the options given
below:

(A) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(B) Both (A) and (R) are true and (R) is the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.

6) The expression given below shows the variation of velocity (n) with time (t), . The
dimension of ABC is :

(A)
(B)
(C)
(D)

7) Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively.
If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the
value of induced emf in coil 1 will be

(A)

(B)

(C)
(D)

8) At the interface between two materials having refractive indices n1 and n2, the critical angle for
reflection of an em wave is θ1C. The n2 material is replaced by another material having refractive
index n3, such that the critical angle at the interface between n1 and n3 materials is θ2C. If n3 > n2 >

n1; and , then θ1C is :

(A)

(B)

(C)

(D) None of these

9) Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The
current is uniformly distributed across this cross-section. The distances from the centre of the wire's
cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum
possible magnetic field, any where due to the wire, will be

(A)

(B)

(C)

(D)

10) As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to
the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre.
Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as

acceleration due to gravity)

(A)

(B)

(C)

(D)
11) Given below are two statements : one is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a
sufficiently negative electron potential to the photoemissive substance.
Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of
a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given
below:

(A) (A) is false but (R) is true.

(B) (A) is true but (R) is false.
(C) Both (A) and (R) are true and (R) is the correct explanation of (A).
(D) Both (A) and (R) are true but (R) is not the correct explanation of (A).

12) A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field
about an axis perpendicular to . Magnetic flux φ and induced emf ε across it, at an instant when
is parallel to the plane of coil, are :

(A) φ = AB, ε = 0
(B) φ = 0, ε = NABω
(C) φ = 0, ε = 0
(D) φ = AB, ε = NABω

13) The fractional compression of water at the depth of 2.5 km below the sea level is
________%. Given, the Bulk modulus of water = 2 × 109 Nm–2, density of water = 103 kg m–3,
acceleration due to gravity = g = 10 ms–2.

(A) 1.75
(B) 1.0
(C) 1.5
(D) 1.25

14) If λ and K are de Broglie Wavelength and kinetic energy, respectively, of a particle with constant
mass, the correct graphical representation for the particle will be :-

(A)
(B)

(C)

(D)

15)
For the circuit shown above, equivalent GATE is :

(A) OR gate
(B) NOT gate
(C) AND gate
(D) NAND gate

16) A body of mass 'm' connected to a massless and unstretchable string goes in verticle circle of
radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top
of circular path is , where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at
top of the circle is:

(A)

(B)

(C)

(D)
17) Let u and v be the distances of the object and the image from a lens of focal length f. The correct
graphical representation of u and v for a convex lens when |u| > f, is

(A)

(B)

(C)

(D)

18) Match List-I with List-II.

List-I List-II

Electric field inside (distance r > 0 from center) of a

(A) uniformly charged spherical shell with surface charge (I) σ / ε0
density σ, and radius R.

Electric field at distance r > 0 from a uniformly charged

(B) (II) σ / 2ε0
infinite plane sheet with surface charge density σ.

Electric field outside (distance r > 0 from center) of a

(C) uniformly charged spherical shell with surface charge (III) 0
density σ, and radius R

Electric field between 2 oppositely charged infinite

(D) plane parallel sheets with uniform surface charge (IV)
density σ.
Choose the correct answer from the options given below :
(A) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(B) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(D) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

19) The workdone in an adiabatic change in an ideal gas depends upon only :

(A) change in its pressure

(B) change in its specific heat
(C) change in its volume
(D) change in its temperature

20) Given below are two statements : one is labelled as Assertion (A) and other is labelled as
Reason (R).
Assertion (A) : Electromagnetic waves carry energy but not momentum.
Reason (R) : Mass of a photon is zero.
In the light of the above statements, choose the most appropriate answer from the options given
below :

(A) (A) is true but (R) is false.

(B) (A) is false but (R) is true.
(C) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(D) Both (A) and (R) are true and (R) is the correct explanation of (A).

SECTION-II

1) The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters.
If a force of acts on the particle, then the magnitude of torque (with respect to origin) in
z-direction is _________.

2) A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the
temperature of gas should be raised to _______ °C.

3) Two light beams fall on a transparent material block at point 1 and 2 with angle θ1 and θ2 ,
respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on
the interface at other end of the block. Given : the distance between 1 and 2, and

, where refractive index of the block n2 > refractive index of the outside
medium n1, then the thickness of the block is ________ cm.

4) In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston
is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the
work done is _________ kJ.

5) The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a
river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s.
Range of the ball as observed by an observer at rest on the river bank, is _________ cm. (Take g = 10
2
m/s )

PART-2 : CHEMISTRY

SECTION-I

1) Total number of nucleophiles from the following is :-

NH3, PhSH, (H3C)2S, H2C=CH2, , H3O⊕, (CH3)2 CO, NCH3

(A) 5
(B) 4
(C) 7
(D) 6

2) The standard reduction potential values of some of the p-block ions are given below. Predict the
one with the strongest oxidising capacity.

(A)

(B)

(C)

(D)
3) The molar conductivity of a weak electrolyte when plotted against the square root of its
concentration, which of the following is expected to be observed?

(A) A small decrease in molar conductivity is observed at infinite dilution.

(B) A small increase in molar conductivity is observed at infinite dilution.
(C) Molar conductivity increases sharply with increase in concentration.
(D) Molar conductivity decreases sharply with increase in concentration.

4) At temperature T, compound AB2(g) dissociates as AB2(g) AB(g) + B2(g) having degree of

dissociation x (small compared to unity). The correct expression for x in terms of Kp and p is

(A)

(B)

(C)

(D)

5) Match List-I with List-II.

List-I (Structure) List-II (IUPAC Name)

(A) (I) 4-Methylpent-1-ene

(B) (CH3)2C (C3H7)2 (II) 3-Ethyl-5-methylheptane

(C) (III) 4,4-Dimethylheptane

(D) (IV) 2-Methyl-1,3-pentadiene

Choose the correct answer from the options given below:

(A) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(B) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(C) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(D) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

6) Choose the correct statements.

(A) Weight of a substance is the amount of matter present in it.
(B) Mass is the force exerted by gravity on an object.
(C) Volume is the amount of space occupied by a substance.
(D) Temperatures below 0ºC are possible in Celsius scale, but in Kelvin scale negative temperature
is not possible.
(E) Precision refers to the closeness of various measurements for the same quantity.
(A) (B), (C) and (D) Only
(B) (A), (B) and (C) Only
(C) (A), (D) and (E) Only
(D) (C), (D) and (E) Only

7) The correct increasing order of stability of the complexes based on Δo value is :

(I) [Mn(CN)6]3– (II) [Co(CN)6]4–
(III) [Fe(CN)6]4– (IV) [Fe(CN)6]3–

(A) II < III < I < IV

(B) IV < III < II < I
(C) I < II < IV < III
(D) III < II < IV < I

8)

Match List-I with List-II.

List-II
List-I
(Hybridisation & Magnetic
(Complex)
characters)

(A) [MnBr4]2– (I) d2sp3 & diamagnetic

(B) [FeF6]3– (II) sp3d2 &paramagnetic

(C) [Co(C2O4)3]3– (III) sp3 & diamagnetic

(D) [Ni(CO)4] (IV) sp3 & paramagnetic

Choose the correct answer from the options given below :
(A) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(B) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(C) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(D) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

9) In the following substitution reaction :

Product ‘P’ formed is :

(A)

(B)

(C)

(D)

10)

For a Mg | Mg2+ (aq) || Ag+(aq) | Ag the correct Nernst Equation is :

(A)

(B)

(C)

(D)

11) The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :

(A) Fe < Mn, Ru < Tc and Re < Os

(B) Mn < Fe, Tc < Ru and Re < Os
(C) Mn < Fe, Tc < Ru and Os < Re
(D) Fe < Mn, Ru < Tc and Os < Re
12) 1.24 g of AX2 (molar mass 124 g mol–1) is dissolved in 1 kg of water to form a solution with
boiling point of 100.0156°C, while 25.4 g of AY2 (molar mass 250 g mol–1) in 2 kg of water constitutes
a solution with a boiling point of 100.0260°C.
Kb(H2O) = 0.52 K kg mol–1
Which of the following is correct ?

(A) AX2 and AY2 (both) are completely unionised.

(B) AX2 and AY2 (both) are fully ionised.
(C) AX2 is completely unionised while AY2 is fully ionised.
(D) AX2 is fully ionised while AY2 is completely unionised.

13) 500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final
temperature and the change in internal energy respectively are :
Given : R = 8.3 J K–1 mol–1

(A) 348 K and 300 J

(B) 378 K and 300 J
(C) 368 K and 500 J
(D) 378 K and 500 J

14) The reaction A2 + B2 → 2 AB follows the mechanism

A + B2 AB + B (slow)
A + B → AB (fast)
The overall order of the reaction is :

(A) 1.5
(B) 3
(C) 2.5
(D) 2

15) If a0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ)
of the electron present in the second orbit of hydrogen atom ? [n : any integer]

(A)

(B)

(C)

(D)

16) The product (P) formed in the following reaction is :

 Product (P)

(A)

(B)

(C)

(D)

17) An element ‘E’ has the ionisation enthalpy value of 374 kJ mol–1. ‘E’ reacts with elements A, B, C
and D with electron gain enthalpy values of –328, –349, –325 and –295 kJ mol–1, respectively.
The correct order of the products EA, EB, EC and ED in terms of ionic character is :

(A) EB > EA > EC > ED

(B) ED > EC > EA > EB
(C) EA > EB > EC > ED
(D) ED > EC > EB > EA

18) Match List – I with List – II.

List – I List – II
(Carbohydrate) (Linkage Source)
 (A) Amylose (I) β-C1-C4, plant

(B) Cellulose (II) α-C1-C4, animal

(C) Glycogen (III) α-C1-C4, α-C1-C6, plant

(D) Amylopectin (IV) α-C1-C4, plant

Choose the correct answer form the options given below :
(A) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(B) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(C) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(D) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

19) The steam volatile compounds among the following are :

(A) (B)

(C) (D)
Choose the correct answer from the options given below :

(A) (B) and (D) only

(B) (A) and (C) only
(C) (A) and (B) only
(D) (A),(B) and (C) only

20) Given below are two statements :

Statement (I) : The radii of isoelectronic species increases in the order.
Mg2+ < Na+ < F¯ < O2–
Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order.
Cl > F > Br > I
In the light of the above statements, choose the most appropriate answer from the options given
below :

(A) Statement I is incorrect but Statement II is correct

(B) Both Statement I and Statement II are incorrect
(C) Statement I is correct but Statement II is incorrect
(D) Both Statement I and Statement II are correct

SECTION-II

1) Given below are some nitrogen containing compounds.

Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume
______mg of HCl.
(Given molar mass in g mol–1 C:12, H : 1, O : 16, Cl : 35.5)

2) The molar mass of the water insoluble product formed from the fusion of chromite ore (FeCr2O4)
with Na2CO3 in presence of O2 is ______g mol–1.

3) The sum of sigma (σ) and pi(π) bonds in Hex-1,3-dien-5-yne is _______.

4) If A2B is 30% ionised in an aqueous solution, then the value of van’t Hoff factor (i) is _____×10–1.

5)
0.1 mole of compound ‘S’ will weigh_____g.
(Given molar mass in g mol–1 C:12, H:1, O:16)

PART-3 : MATHEMATICS

SECTION-I

1) Let the line x + y = 1 meet the circle x2 + y2 = 4 at the points A and B. If the line perpendicular to
AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area
of the quadrilateral ADBC is equal to

(A)
(B)
(C)
(D)

2) Let M and m respectively be the maximum and the minimum values of

Then M4 – m4 is equal to :

(A) 1280
(B) 1295
(C) 1040
(D) 1215

3)

Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis,
respectively. If these parabolas intersects at the points A and B, then (AB)2 is equal to

(A) 192
(B) 384
(C) 96
(D) 392

4) Let ABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y –31 = 0 and 9x –2y–19 = 0, Let
the point (h,k) be the image of the centroid of ΔABC in the line 3x + 6y –53 = 0. Then h2 + k2 + hk is
equal to

(A) 37
(B) 47
(C) 40
(D) 36

5) Let , and be a vector such that and

= 168. Then the maximum value of is :

(A) 77
(B) 462
(C) 308
(D) 154

6) Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P
are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is :

(A) 158
(B) 173
(C) 164
(D) 161
7) Let the area of the region { , , }be A. Then 6A is equal to :

(A) 16
(B) 12
(C) 18
(D) 14

8) The least value of n for which the number of integral terms in the Binomial expansion of

is 183, is :

(A) 2184
(B) 2148
(C) 2172
(D) 2196

9) The number of solutions of the equation is :

(A) 2
(B) 4
(C) 1
(D) 3

10) Let y = y(x) be the solution of the differential equation cosx(loge(cosx))2dy + (sinx –3ysinx loge

(cosx))dx = 0, . If then is :

(A)

(B)

(C)

(D)

11) Define a relation R on the interval by x R y if and only if sec2x – tan2y = 1. Then R is :

(A) an equivalence relation

(B) both reflexive and transitive but not symmetric
(C) both reflexive and symmetric but not transitive
(D) reflexive but neither symmetric not transitive
12) Let the ellipse, , a > b and A < B have same eccentricity

Let the product of their lengths of latus rectums be , and the distance between the foci of E1 be
4. If E1 and E2 meet at A,B,C and D, then the area of the quadrilateral ABCD equals :

(A)

(B)

(C)

(D)

13) Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the
first twenty terms lies between 1600 and 1800. Then its 11th term is :

(A) 84
(B) 122
(C) 90
(D) 108

14) Let and . Let L1: , λ ∈ R and L2 :

, µ ∈ R, be two lines. If the line L3 passes through the point of intersection of L1 and
L2, and is parallel to , then L3 passes through the point:

(A) (8, 26, 12)

(B) (2, 8, 5)
(C) (–1, –1, 1)
(D) (5, 17, 4)

15) The value of is :

(A)

(B) 2

(C)

(D)

16) The integral dθ is equal to :

(A) 3 loge4
(B) 6 loge4
(C) 4 loge3
(D) 2 loge3

17) Let and be two lines. Let L3 be a line passing

through the point (α,β,γ) and be perpendicular to both L1 and L2. If L3 intersects L1, then
|5α–11β–8γ| equals :

(A) 18
(B) 16
(C) 25
(D) 20

18) Let x1, x2,......x10 be ten observations such that , , β > 2 and

their variance is . If μ and σ2 are respectively the mean and the variance of 2(x1–1) + 4β, 2(x2–1) +

4β, ....., 2(x10 – 1) + 4β, then is equal to :

(A) 100
(B) 110
(C) 120
(D) 90

19) Let |z1 – 8 + 2i| ≤ 1 and |z2 – 2 + 6i| ≤ 2, z1, z2 ∈ C. Then the minimum value of |z1 – z2| is :

(A) 3
(B) 7
(C) 13
(D) 10

20) Let , If Aij is the cofactor of aij, Cij = , , and C

= [Cij], then 8|C| is equal to :

(A) 262
(B) 288
(C) 242
(D) 222

SECTION-II
1) Let f : (0,∞) → R be a twice differentiable function. If for some a ≠ 0, = af(x), f(1) = 1

and f(16) = , then 16 – is equal to ______,

2) Let , where . Then n(S) is equal to ____.

3) Let [t] be the greatest integer less than or equal to t. Then the least value of for which

is equal to ____.

4) The number of 6-letter words, with or without meaning, that can be formed using the letters of
the word MATHS such that any letter that appears in the word must appear at least twice, is 4 ____.

5) Let . Then is equal to ______.

 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. C C D B B A C D B A D B D B A D B D D B

SECTION-II

Q. 21 22 23 24 25
A. 2 327 6 5 2000

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. A D D C C D C D A B C D A A B C A B C D

SECTION-II

Q. 46 47 48 49 50
A. 341 160 15 16 13

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. B A A A C D D A B D A D C A D C C A B C

SECTION-II

Q. 71 72 73 74 75
A. 112 2 24 1405 5
 SOLUTIONS

PART-1 : PHYSICS

1)

A: Correct
B : Correct with correct explanation

2)

3) Iω2θ = qℓE0θ

4)

[τ] ≈ [E]

[σ] ≠ [I]
[L] = [h]
[P] = [Y]

5) As h increases, g decreases, T increases

6) [LT–1] = [A] [T2] =

[C] = [T]
[A] = [LT–3]
[B] = [LT–1]
[ABC] = [L2T–3]

7) ϕ1 = L1 I1 + M12 I2

8) Ans. Dropped by NTA

9) Maximum possible magnetic field is at the surface

It can be obtained inside as well as outside the wire For inside,

For outside

⇒ r = 2a

Correct answer

10)
Velocity of a just before hitting :
Just after collision, let velocity of A and B are v1 and v2 respectively
∴ by COM:

mu = mv1 +
2v1 + v2 = 2u ….(i)

⇒ v2 – v1 = u ….(ii)
From (i) –(ii)

⇒ 3v1 = u

11) (A) : True

(B) : True but not correct explanation

12)
ϕ = BAN.cos(ωt)

= BAωN.sin(ωt)

When B is parallel to plane, ωt =

⇒ ϕ = 0, ε = BAωN

13)

= 1.25 %

14)

Y = cx2
Upward facing parabola passing through origin.
15)
⇒ OR Gate

16)

17) (u + f)(v – f) = f2

18) (A) → 0 (III)

(B) → (II)

(C) → (No row matching)

(D) → (I)

19) ΔW = –ΔU = – nCVΔT

20) Assertion is false because em waves have momentum.

21)
22)

T2 = 600 K
T2 = 327°C

23)
n1sin(90 – θ1) = n2sinθ3
n1cosθ1 = n2sinθ3

, θ3 = 30

24)

, ,
F = 50 × 500 = 25 ×103 N

= 5 × 103 = 5 kJ

25) = 9 + 27 = 36 km/hr

= 10 m/sec

Time of flight = = 2 sec

Range = 10 × 2 = 20m = 2000 cm
PART-2 : CHEMISTRY

26) Total five nucleophiles are present

NH3, PhSH, (H3C)2S, CH2=CH2,

27) Standard reduction potential value (+ve) increases oxidising capacity increases.

28)

29) AB2(g) AB(g) + B2(g)

teq.

⇒ x << 1 ⇒ 1and 1 – x 1

30) (A)
3-Ethyl-5-methylheptane
(B) (CH3)2C (C3H7)2

4, 4-Dimethylheptane

(C)
2-Methyl-1, 3-pentadiene

(D)
4-Methylpent-1-ene
31) Theory based

32)

(I) [Mn(CN)6]3– –1.6 Δo

(II) [Co(CN)6]4– –1.8 Δo
(III) [Fe(CN)6]4– –2.4 Δo
(IV) [Fe(CN)6]3– –2.0 Δo
I < II < IV < III

33) (A) [MnBr4]2–

Mn+2 ⇒ [Ar] 3d5
In presence of ligand field

⇒ [Ar]
⇒ sp3 hybridization, paramagnetic in nature
3–
(B) [FeF6]
Fe+3 ⇒ [Ar] 3d5
In presence of ligand field

⇒ [Ar]
⇒ sp3d2 hybridization, paramagnetic in nature
(C) [Co(C2O4)3]3–
Co+3 ⇒ [Ar] 3d6
In presence of ligand field

⇒ [Ar]
⇒ d2sp3 hybridization, diamagnetic in nature
(D) [Ni(CO)4]
0
Ni ⇒ [Ar] 3d84s2
In presence of ligand field

⇒ [Ar]
⇒ sp3 hybridization, diamagnetic in nature

34) It is an example of nucleophillic Aromatic substitution reaction.

35) According to Nernst equation :-

E = E° – ln Q.
Cell reaction :-
Mg(s) + 2Ag(s) +

⇒Q=

⇒E= – ln

36) M.P. ⇒ Mn < Fe, Tc < Ru, Os < Re

NCERT based

37) For AX2 :- ΔTb = Kb × m × i

0.0156 = 0.52 × ×
⇒ = 3 ⇒ complete ionisation
For AY2 :- ΔTb = Kb × m × i
0.026 = 0.52 × 0.0508 ×
⇒ 1 ∴ complete unionisation

38) qp = n × cp × ΔT

⇒ 500 = 0.5 × × 8.3 (Tf –298)

⇒ Tf 346.2K

= =

⇒ ΔU = × 500 = 300 J

39) rate = k2[A][B2] …….(1)

⇒ [A] =
Substituting in (1) ; we get

Rate = k2 . . [B2]

∴ order = = 1.5

40) 2πrn = nλ
2π .(4a0) = nλ

=λ=

41)

42) Difference between I.E. & E.G.E increases, ionic character increases.

43) Informative

44) (A) & (B)

are steam volatile due to intramolecular hydrogen bonding.

45) (i) For isoelectronic species –ve charge increases, radii increases.
(ii) Magnitude of E.G.E : Cl > F > Br > I

46) Benzyl Amine is most basic due to localised lone pair.

Mole of benzyl Amine ⇒ mole

1 Mole of Benzyl amine consumed 1 mole of HCl
So, Mole of HCl consumed → 0.00934 mole
Mass of HCl consumed → 0.00934 × molar mass of HCl
= 0.00934 × 36.5
= 0.341 gm
= 341 mg

47) 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2

Fe2O3 is water insoluble, so its molar mass
⇒ [2 × 56 + 3 × 16] ⇒ 160 g/mol
 48)
Number of σ bond = 11
Number of π bond = 4
σ + π = 11 + 4 = 15

49) A2B → 2A+ + B–2 ; y = 3

α = 0.3
i = 1 + (y – 1)α
= 1 + (3 – 1) (0.3) = 1.6 = 16 × 10–1

50)
0.1 mole of compound (S) weight in gm
= 0.1 × molar mass of compound (S)
= 0.1 × 130 = 13 gm

PART-3 : MATHEMATICS

51)
By solving x = y with circle
We get

By solving x + y = 1 with
circle x2 + y2 = 4
we get

&
∴ Area of Quadrilateral ACBD
= 2 × Area of ΔBCD
=
=

52)
R2 → R2–R1 & R3 → R3 – R1

Expand about R1 , we get

f(x) = 2 + 4sin4x
∴ M = max value of f(x) = 6
m = min value of f(x) = –2
∴ M4 – m4 = 1280

53)
Let intersection points of these two parabolas are A(x1, y1) & B(x2,y2)
∵ equation of parabola I and II are given below
∴ (x – 4)2 + (y –3)2 = x2 …..(1)
& (x – 4)2 + (y – 3)2 = y 2 …..(2)
Here A(x1, y1) & B(x2,y2) will satisfy the equation
Also from equations (1) & (2), we get x = y ..(3)
Put x = y in equation (1)
We get x2 – 14x + 25 = 0
x1 + x2 = 14
x1x2 = 25
∴ AB2 = (x1 – x2)2 + (y1–y2)2
= 2(x1 – x2)2
= 2[(x1+x2)2 – 4x1x2] = 192
54)

∴ centroid of ΔABC =

Let image of centroid with respect to line mirror is (h,k)

∴ &
Solving (1) & (2) we get h = 3, k = 4
∴ h2 + k2 + hk = 37

55)

=0

⇒
....(1)

using equation (1)

+ 77λ2 = 154
77λ + 77λ2 – 154 = 0
λ2 + λ – 2 = 0
λ = –2, 1
∴ Maximum value of occurs when λ = –2

= 77×4
= 308

56) (i) number of numbers created using

1111133 = ⇒ 21
(ii) number of numbers created using

1111223 = ⇒ 105
(iii) number of numbers created using

1112222 = ⇒ 35
Total = 161

57)
A ⇒ Rectangle ABDE – Area of region EDC

A ⇒ 4 –2

A⇒4–2

A⇒4–2 =
So 6A = 14

58) General term =

=
For integral terms, r must be multiple of 12
∴ r = 12k, k ∈ W
Total values of r = 183
Hence max r = 12(182)
= 2184
Min value of n = 2184

59) Consider
(9α2 – 9α + 2)(2α2 – 7α + 3) = 0
(3α – 2)(3α – 1)(α – 3) (2α – 1) = 0

So, no. of solutions = 4

60)
Given : ,

⇒C=0

∴ y (ln(secx))3 = (ln(secx))2 + 0

y=

y=

∴y =

61) sec2x – tan2x = 1 (on replacing y with x)

⇒ Reflexive
sec2x – tan2y = 1
⇒ 1 + tan2x + 1 – sec2y = 1
⇒ sec2y – tan2x = 1
⇒ symmetric
sec2x – tan2y = 1, sec2y – tan2z = 1
Adding both
⇒ sec2x – tan2y + sec2y – tan2 z = 1 + 1
sec2x + 1 – tan2z = 2
sec2x – tan2z = 1
⇒ Transitive
hence equivalence relation
62) 2ae = 4

Now ⇒
⇒ A2 = 2B

⇒ ⇒B=3
⇒ A2 = 6

…..(1)

…..(2)
On solving (1) & (2) we get

(x, y) ≡

The four points are vertices of rectangle and its area =

63) S3 = 3a + 3d = 54
⇒ a + d = 18
S20 = 10(2a + 19d)
⇒ 10(36 + 17d)
⇒ 1600 < 10(36 + 17d) < 1800
⇒ 160 < 36 + 17d < 180
⇒ 124 < 17d < 144

⇒ <d<
Common difference will be natural number
⇒ d = 8 ⇒ a = 10
⇒ a11 = 10 + 10 × 8 = 90

64) L1 : =
=
L2 : =
⇒ =
For point of intersection equating respective components
⇒ λ – 1 = 2μ ...(1)
2(λ + 1) = 1 + 7μ ....(2)
λ + 1 = 1 + 3μ .....(3)
We get
⇒ λ = 3 and μ = 1
⇒
L3 : =
For α = 2, =

65)

66)

Let sinθ – cosθ = t

(cosθ + sinθ)dθ = dt

= 2ln(1) + 4ln3
= 4ln3
67) DR’s of L3 = =

L3 :
A(α – 5λ, β – 3λ, γ + λ)

L1 :
B(k + 1, –k + 2, 2k + 1)
Now
α –5λ = k+1 ⇒ α = 5λ + k +1
β – 3λ = –k+2 ⇒ β = 3λ – k + 2
γ + λ = 2k + 1 ⇒ γ = –λ + 2k +1
|5α –11β – 8γ| = |–25|
= 25

68) , ∴ mean = 5

Variance =

⇒ = 258 ....(1)

Now

258 –2β(50) + 10β2 = 98

(β – 8)(β – 2) = 0
β = 8 or β = 2 (as β > 2)
∴β=8 ....(2)
Now as per the question
2(x1–1) + 4β, 2(x2–1) + 4β,….2(x10–1)+4β
can be simplified to
2x1 + 30, 2x2 + 30,.....2x10 + 30 using eq. (2)
μ = 2(5) + 30 = 40 ....(3)

σ2 = 22 =
69)
∵ AB =
∴ |Z1 – Z2|min = 10 – 2 – 1 = 7

70) |A| =

C11 =

C12 =

C21 =

C22 =

C=

|C| =
8|C| = 242

71)
λx = t

dλ =

f(x) = a(x f’(x) + f(x))

(1 – a)f(x) = a.x f’(x)
ℓnf(x) = ℓnx + c
x = l, f(1) = l ⇒ c = 0

x = l6, f(16) =

⇒–3= ⇒a=4
f(x) =

f’(x) =

∴ 16 – f’

=
= 16 + 96 = 112

72)

and so on

=
= m2 + 1 + m + 1 = 8
= m2 + m–6 = 0 ⇒ m = –3, 2
n(s) = 2

73) 1
(1 + 2 + ......+ p) – (12 + 22 + ....92) ≥ 1
p (p + 1) ≥ 572
Least natural value of p is 24

74) (i) Single letter is used , then no. of words = 5

(ii) Two distinct letters are used, then no. of words

(iii) Three distinct letters are used, then no. of words

Total no. of words = 1405

75) cos–1x = π + sin–1x + sin–1 (2x + 1)

2cos–1x–sin–1(2x + 1) =

–1 –1
2α – β = where cos x = α, sin (2x+1) = β

2α = +β
cos2α = sinβ
2cos2α –1 = sinβ
2x2 –1 = 2x +1
x2 – x –1 = 0

⇒x=
∴ 4x2 –4x = 4
(2x – 1)2 = 5