10-08-2025

1001CJA101022250024 JA

PART-1 : PHYSICS

SECTION-I

1) Equivalent resistance between A & B is :-

(A) 1Ω
(B) 36Ω
(C) 3Ω
(D) 12Ω

2) In the given circuit find the magnitude of current flowing through arm 'AB' :

(A) 4.5 A
(B) 1.5 A
(C) 3.5 A
(D) 4 A

3) Statement-1 : The total energy of the satellite is always negative irrespective of the nature of its
orbit, i.e. elliptical or circular and it cannot be positive or zero.
Statement-2 : If the total energy is negative the satellite would leave its orbit.

(A) Statement-1 is true, Statement-2 is true. Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true. Statement-2 is NOT the correct explanation of
(B)
Statement-1.
(C) Statement-1 is true, Statement-2 is false.
(D) Statement-1 is false, Statement-2 is ture.

4) A galvanometer of resistance 20 Ω gives a full scale deflection when a current of 0.04 A is passed
through it. It is desired to convert it into an ammeter of full scale reading 20 A. The only shunt
available is 0.050 Ω resistance. The resistance that must be connected in series with the coil of the
galvanometer is :-

(A) 4.95 Ω
(B) 5.94 Ω
(C) 0.01 Ω
(D) 0.04 Ω

5) The circuit shown below shows a steady state with S open. When S is closed :-

(A) No charge/current passes through S.

(B) A steady current of 30 A flows through S.
(C) 90µC of charge flows from B to A.
(D) 120 µC of charge flows from A to B.

6) Five identical parallel conducting plates each of area A have separation ‘d’ between successive
surfaces. The plates are connected to the terminal of a battery as shown in the fig. The effective

capacitance of the circuit is :-

(A)

(B)

(C)

(D)

7) A light ray is incident on a prism of prism angle 60° and refractive index as shown in figure.
The angle of emergence e is given by :

(A) 30°
(B) 45°
(C) 90°
(D) 60°

8) Two plane mirrors are inclined at an angle of 72°. An object is kept between them symmetrically,

then number of images of object is

(A) 4
(B) 5
(C) 6
(D) 3

9)

Match the entries on left side with entries on right side when left most plate is grounded. Then

List-I List-II
(P) The charge appearing on outer surface of right most plate is (1) 0
(Q) The charge appearing on outer surface of left most plate is (2) Q
(R) The charge appearing on left surface of middle plate is (3) –2Q
(S) The charge appearing on right surface of middle plate is (4) 2Q
(A) (P) - 1; (Q) - 2; (R) - 3; (S) - 3
(B) (P) - 1; (Q) - 1; (R) - 3; (S) - 2
(C) (P) - 1; (Q) - 1; (R) - 3; (S) - 4
(D) (P) - 2; (Q) - 2; (R) - 3; (S) - 4

10) Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular
deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is :-

(A) 40Ω
(B) 25Ω
(C) 250Ω
(D) 500Ω

11) A charge particle moves through a uniform magnetic field. The direction of initial velocity is
perpendicular to the magnetic field. Then mark the CORRECT option.

(A) Kinetic energy changes but momentum remains constant.

(B) Momentum changes but kinetic energy remains constant.
(C) Both momentum & kinetic energy changes.
(D) Both momentum & kinetic energy remains constant.

12) In the figure, there is a conducting wire having current i and which has a shape of half ellipse

is kept in a uniform magnetic field B as shown. If the mass of wire is m, the

acceleration of wire will be :-

(A)

(B)

(C)

(D)

13) A graph between velocity and displacement of a particle performing SHM is shown. Find the
acceleration when the particle is at half of maximum displacement from mean position.
(A) 10 cm/s2
(B) 20 cm/s2
(C) 30 cm/s2
(D) 40 cm/s2

14) The potential energy of a particle of mass 0.1 kg, moving along x-axis under the action of a
single force, is given by U = 5x (x – 4) J where x is in meters. Choose the INCORRECT statement.

(A)
The period of oscillation of the particle is .
(B) The speed of the particle is maximum at x = 2m.
(C) The particle executes simple harmonic motion.
(D) The particle is acted upon by a constant force.

15) Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys
Maxwell's distribution.
Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the
translational kinetic energy for each molecule.

(A) Statement I is false but Statement II is true.

(B) Both Statement I and Statement II are false.
(C) Both Statement I and Statement II are true.
(D) Statement I is true but Statement II is false.

16) If the conducting rod accelerates with acceleration along rod length and vA and vB are potential

of end A and B respectively, then :-

(A) Potential of point A is equal to that of B.

(B) Potential of point B is less than that of A.
(C) A constant electric field appears in the rod from B to A.
(D) A constant electric field appears in the rod from A to B.

17) The radius (in m) of the spherical surface of zero potential due to charges +2q and –3q fixed
at (4, 0, 0)m and (9, 0, 0)m respectively, is :- (Assume V = 0 at infinity)

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

18) A spherical ball of transparent material has a refractive index µ. A narrow beam of light from air
is aimed as shown. The value of refractive index so that light is focused at point C on the opposite

end of the diameter, is :

(A)

(B)

(C)

(D) 2

19) A plane wave is described by the equation y = 3 cos m. The maximum velocity (in
m/s) of the particles of the medium due to this wave is

(A) 20
(B) 30
(C) 40
(D) 50

20)

The square loop has sides of length 2cm. A magnetic field points out of the page and its magnitude is
given by B = (4t2y) T. The emf induced in loop (in µV) at t = 2.5 s is :-

(A) 60
(B) 70
(C) 80
(D) 90
 SECTION-II

1) Consider a metallic ring of radius 1 m, mass 1 kg and carrying a current of 1A in a gravity free
space in the x-y plane with its centre O at the origin as shown in the figure. If a uniform magnetic
field T is applied, then the instantaneous acceleration of the point P (which is on the y-axis at

the moment) will be C × π m/s2. Find the value of C.

2) An alternating voltage V0 = 100 V with angular frequency ω is connected across the capacitor and
inductor having XL = 5Ω and XC = 10Ω. Find the ratio of current through inductor to AC source.

3) A uniform magnetic field of 0.06 T is inside the plane of the figure. The resistance of rod is 25Ω,
mass is 36 gm and it can slide freely on smooth parallel rails which are perfectly conducting. The
whole system is in horizontal plane. The rod starts from rest. If terminal velocity of the rod (in m/s) is

v then find .

4) One mole of a monoatomic gas is heated in such a way that its molar specific heat is 2R. During
the heating, the volume of the gas is doubled. By what factor does the temperature of the gas
change?

5) The adiabatic container shown in figure has two chambers, separated by a partition, of volumes V1
= 2.0 litre and V2 = 3.0 litre. The chambers contain μ1 = 4.0 and μ2 = 5.0 moles of a gas at pressures
p1 = 1.00 atm and p2 = 2.00 atm. The pressure after the partition is removed and the mixture attains

equilibrium is P atm, then the value of , Find N.

 PART-2 : CHEMISTRY

SECTION-I

1)

For three complex compounds

correct information is :-

(A) Order of Δ0 will be I > II > III

(B) Order of λabs will be III > II > I
(C) All complex will show optical isomerism
(D) All complex will not show geometrical isomerism

2) Consider the following complexes.

[PtCl4]2– , [Pt(CN)4]2– , [Ni(CO)4]
What is/are the similarity in all the three complexes :

(A) Magnetic nature

(B) Oxidation state of central metal
(C) Geometry and hybridisation
(D) Maximum number of atom(s) in a plane

3) Which one of the following statement is not true regarding diborane?

It has two bridging hydrogen and four hydrogen perpendicular with respect to bridging
(A)
hydrogen
(B) When methylated, the product is Me4B2H2
(C) sp3 - s - sp3 type of overlapping is present
(D) All the B–H bond distances are equal.

4) The IUPAC name of [Ni(NH3)4] [NiCl4] is -

(A) Tetrachloronickel(II) - tetraamminenickel(II)

(B) Tetraamminenickel(II) - tetrachloronickel (II)
(C) Tetraamminenickel(II) - tetrachloronickelate(II)
(D) Tetrachloronickel(II) - tetraamminenickelate(0)

5) Which of the following statement is incorrect

(A) Hall's process is used for leaching of Red Bauxite.

(B) In Mac Arthur forrest cyanide process Zn dust is used as Reducing agent
(C) According to MOT, O2 is paramagnetic while is diamagnetic
(D) Serpeck process is used for leaching of white Bauxite

6) From following table, Identify correct matching process involves in Extraction

Column-I Column-II

P NaCl 1 Roasting

Q Zinc blende 2 Calcination

R Dolomite 3 Electrolytic Reduction

S Pbo 4 Carbon Reduction

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

7)
Major organic product will be-

(A)

(B)

(C)

(D)
8) The pH of the solution produced when an aqueous solution of strong acid pH 5 is mixed with
equal volume of an aqueous solution of strong acid of pH 3 is nearly :-

(A) 3.3
(B) 4.3
(C) 5.3
(D) 6.3

9) HX is a weak acid (Ka = 10–5). It forms a salt NaX (0.1 M) on reacting with caustic soda. The
degree of hydrolysis of NaX is

(A) 0.01%
(B) 0.0001%
(C) 0.1%
(D) 0.5%

10) For the following Assertion and Reason, the correct option is :
Assertion : The pH of water increases with increase in temperature.
Reason : The dissociation of water into H+ and OH– is an exothermic reaction.

Both assertion and reason are true, but the reason is not the correct explanation for the
(A)
assertion.
(B) Both assertion and reason are false.
(C) Assertion is not true, but reason is true.
(D) Both assertion and reason are true, and the reason is the correct explanation for the assertion.

11) For a reaction C(graphite) + 2H2(g) —→ CH4(g), ΔH° = –75 KJ at 300K and 1 atm. The standard

entropies of C(graphite), H2(g) and CH4(g) are 6, 130.6 and 186.2 respectively. for
the reaction will be :-

(A) –81
(B) 169
(C) –169
(D) 81

12) An ideal gas heat engine operates by carnot's cycle between 227ºC and 127ºC. It absorbs 6 ×
104 cal of heat at high temperature. Amount of heat converted to work is :

(A) 2.4 × 104 cal

(B) 4.8 × 104 cal
(C) 1.2 × 104 cal
(D) 6.0 × 104 cal

13) The heats of formation of CO2(g) and H2O(l) are –394 kJ/mole and –285.8 kJ/mole respectively
Using the data for the following combustion reaction, calculate the heat of formation of C2H2(g).
2C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) ; ΔHo = – 2601 kJ

(A) – 238.6 kJ/mole

(B) 253.2 kJ/mole
(C) 238.7 kJ/mole
(D) 226.7 kJ/mole

14) Select the correct statement -

(A) ΔHatomisation of graphite is equal to ΔHf [C(g)]

(B) ΔHcomb of ‘H’ atom is equal to ΔHf [H2O(l)]
(C) ΔHf [H2O(l)] is zero
(D) ΔHcomb of graphite is equal ΔHf [CO(g)]

15) Which of the following will give silver mirror image test with Tollen's reagent?

(A)

(B) Glucose

(C)

(D) All of these

16) Epoxide can be synthesized by treating halohydrins with aqueous base, select in which no
epoxide formation is observed-

(A)

(B)

(C)

(D) None of these

17) Major product (Having carbon chain greater
than 5 carbon)

Major product have-

(A) Ether functional group present

(B) Ketal functional group present
(C) Ketone functional group present
(D) Both ketal and ketone functional group

18)
X will be-

(A)

(B)

(C)

(D)
19)
Major product will be-

(A)

(B)

(C)

(D) Both (A) and (B)

20) Statement-I : Chloral form hydrogen bond in hydrated form.

Statement-II : Keq > 1

(A) Both Statement I and Statement II are incorrect.

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

SECTION-II

1) Find total number of ores which are concentrated by Froth Flotation process.
Chromite, Galena, Copper pyrites, Cassiterite, Carnalite, Zinc blende, Dolomite

2) Calculate the number of electrons of 'Na' for which |m| is 1.

3) When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1
M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The
strength of KMnO4 in grams per litre is ________ × 10–2. (Nearest integer)
[Atomic mass of K = 39, Mn = 55, O = 16]

4) The closed cylinder shown in figure has a freely moving piston separating chambers 1 and 2.
Chamber 1 contains 280 mg of N2 gas, and chamber 2 contains 200 mg of helium gas. When
equilibrium is established, what will be the ratio L2/L1 ? (Molar masses of N2 and He are 28 and 4
g/mol respectively).

5) Major organic product

Number of π bonds in major organic product.

PART-3 : MATHEMATICS

SECTION-I

1) If the shortest distance of the parabola y2 = 4x from the centre of the circle x2 + y2 – 4x – 16y + 64
= 0 is d, then d2 is equal to :

(A) 16
(B) 24
(C) 20
(D) 36

2) Let β be a real number. Consider the matrix A = . If A7 – (β – 1)A6 – βA5 is a singular

matrix, then the value of 9β is

(A) 3
(B) 5
(C) 4
(D) 1

3) Assertion (A) : If normal at the ends of double ordinate x = 4 of parabola y2 = 4x meet the curve
again at P and P′ respectively, then PP′ = 12 unit.
Reason (R) : If normal at t1 of y2 = 4ax meets the parabola again at t2, then t12 = 2 + t1 t2.

(A) Both A and R are correct and R is the correct explanation of A

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

4) If y = f(x) be the solution of the differential equation and f(1) = 0, then

f(0) is

(A)

(B)

(C)

(D)

5) Let , tr(M) = α and det(M) = –β, where α = α (θ) and β = β (θ) are
real number, and
α* is the minimum of the set {α(θ) : θ ∈ [0, 2π)} and
β* is the minimum of the set {β (θ): θ∈ [0, 2π)},
then the value of α* + β* is (where tr(M) = trace of matrix M)

(A)

(B)

(C)

(D)

6) Differential equation = x3 cos2 y is represented by family of curves which is given by

: (where C is arbitrary constant)

(A) x6 + 6x2 = C tan y

(B) 6x2 tan y = x6 + C
(C) sin 2y = x3 cos2 y + C
(D) cos 2y = x3 sin2 y + C

7) The curve which satisfy differential equation (x2y – y)dx + (xy2 – x)dy = 0 is-

(A) πx2y2 =
(B) πx2y2 =
(C) exy = ex+y
 exy =
(D)

8) Each side of a square is increasing at the uniform rate of 1 m/sec. If after some time the area of
the square is increasing at the rate of 8 m2/sec, then the area of square at that time in sq. meters is

(A) 4
(B) 9
(C) 16
(D) 25

9) Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II

Degree of differential equation

(A) (I) 0
is

(B) (II) 2
Value of

(C) (III) 1
equal to

(D) (IV) 4
Value of
(A) (A) – II, (B) – I, (C) – IV, (D) – III
(B) (A) – III, (B) – II, (C) – IV, (D) – I
(C) (A) – III, (B) – I, (C) – II, (D) – III
(D) (A) – III, (B) – I, (C) – II, (D) – II

10) If 14a + 9b + 6c = 0 then the equation ax2 + bx + c = 0 has atleast one root in

(A)
(B)
(C)

(D)

11) If I1 = , I2 = , I3 = and I4 = then

(A) I3 > I4
(B) I3 = I4
(C) I1 > I2
(D) I2 > I1

12) If then the value of I is

(A)

(B) 0

(C)

(D)

13) If set of all possible values of p for which the point ([p + 1], 2) lying inside the circle x2 + y2 – x –

y – 8 = 0 is [a, b), (where [.] denotes greatest integer function), then is

equal to -

(A) 2
(B) 0
(C) 4
(D) 6

14) The value of is :

(A)

(B)

(C)

(D)

15) For the function f(x) = (cosx) – x + 1, x ∈ ,

Statement-I : f(x) = 0 for only one value of x in [0, π].

Statement-II : f(x) is decreasing in and increasing in .

(A) Both Statement-I and Statement-II are true

(B) Statement-I is true but Statement-II is false
(C) Both Statement-I and Statement-II are false
(D) Statement-I is false but Statement-II is true

16)

The number of points where

ƒ(x) = (x2 – x) |x3 – 9x2 + 20x| is not differentiable, is-

(A) 2
(B) 3
(C) 0
(D) 4

17) Let f(x) be a continuous function in [–1, 1] such that , then

(p + q + r) is equal to

(A) 3
(B) 2
(C) 1/3
(D) 5

18) Let f : [–1, 3] → [–37, 27] be a function defined by f(x) = 2x3 – 6x2 – 18x + 17. Then f(x) is

(A) injective but not surjective

(B) surjective but not injective
(C) neither injective nor surjective
(D) both injective as well as surjective

19) If S is the sum of the first 10 terms of the series

then tan(S) is equal to

(A)

(B)

(C)

(D)

20) The eccentricity of the conic (2x – 4)2 + 4y2 = (x + y + 1)2 is

(A)

(B)

(C) 2
(D)

SECTION-II

1) Let A = , B = [a b c] and C = be three given matrices,

(where a, b, c and x ∈ ), Given that tr(AB) = tr(C) ∀ x ∈ , (where tr(A) denotes trace of A). If

= (where p, q are coprime), then the value of (p + q) is equal to

2) The slope of tangent to the curve at the point of inflection is

3) Let y = p(x) be the parabola whose axis is parallel to y-axis and passing through the points (–1, 0),

(0, 1) and (1, 0). If the area of the region is A, then 12(π – 4A)
is equal to

4) Area of triangle formed by common tangents to parabola y2 = 8x and hyperbola and

directrix of parabola is Δ then (2Δ) is

5) , 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. A D C A C B D A C C B D B D D C C D B C

SECTION-II

Q. 21 22 23 24 25
A. 8 2 4 4 8

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. D A D C C B C A A B B C D A D C C C A B

SECTION-II

Q. 46 47 48 49 50
A. 3 4 316 5 4

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. C A C B B B A C C A C C A B B A B D D B

SECTION-II

Q. 71 72 73 74 75
A. 29 1 16 9 24
 SOLUTIONS

PART-1 : PHYSICS

1)

All resistances are in parallel ⇒ Req = = 1Ω

2)

Current in branch AB 4 A

3)

If energy of satellite is negative, its orbit is bounded i.e. it remains orbital.

4)
19.96 × 0.05 = 0.04 (20 + R1)
R1 = 4.95 Ω

5)

Charge initially

Charge finally

for this charge arrangement to happen, 90μC charge has to flow upwards.
6)
C = 4C
eq

7)

r1 = 30°
r2 = A – r1
= 30

8)

9)

As potential of left most plate is zero. By using Gauss law we get above options.

10)

Let N = number of turns in galvanometer

A = Area, B = magnetic field
k = the restoring torque per unit twist

Current sensitivity,

Voltage sensitivity,
So, resistance of galvanometer
11) Speed remains constant but direction changes in circular path.

12) Clearly F = BIl = BI × (2b)

∴a=

13)

Speed

10 = w×(2.5)
w=4

14) Question Explanation:

Potential energy of a particle as a function of x is given along with it's mass. Four statements
are given and we have to select the incorrect statement.
Concept:
Potential energy, restoring force and calculation of time period.
Solution:
m = 0.1 kg
u = 5x(x – 4) ⇒ u = 5x2 – 20x

F = –10x + 20 ⇒ F = –10(x – 2)
F = –10(x – 2) k = 10
F = 0 at n = 2 ⇒ n = 2 is mean position.
→ Since, F ∝ –x ⇒ motion of particle is SHM.
→ Velocity of particle is SHM is maximum at mean position.

→ Time period

seconds
→ Since force is varying with x, it is not constant.

15)

Translational degree of freedom = 3

Rotational degree of freedom = 2

16) Due to pseudo force, electrons move from B to A.

17)

18 – 2x = 3x – 12
30 = 5x
x=6

⇒ 10 + 2y = 3y
y = 10
dia = 12 m

r=6m

18)

µ=2

19) vmax = aw = 3 × 10 = 30

20) dϕ = 4t2yℓdy

ϕ=

e=

21)
t = Iα

1 · π (1)2·5 = 1 ·
 ⇒ α = 10π
a = α = (1·sin53) α = 8π

22)
If V = V0 sin ωt

ic =
ic = 10 cos ωt

iL =
iL = – 20 cos ωt
i = ic + iL = –10 cos ωt
So required ratio = 2

23)

ε = Blv = 36

v= = 1200 m/s

24)
∴ k = –1
⇒ PV–1 = constant

= constant ∴ T ∝ V2

25)

neq = n1 + n2

PART-2 : CHEMISTRY
 26)

(A) order of Δo - II > I > III

(B) Xobsorbed - III > I > II
(C) Only [M(en)3]3+ show optical isomerism.
(D) M(AA)3, Ma6 do not show GI.

27)

Complex O.S. Hybridisation Magnetic Nature Geometry C.No.

[PtCl4]2– +2 dsp2 Diamagnetic sq. planar 4

[Pt(CN)4]2– +2 dsp2 Diamagnetic sq. planar 4

[Ni(CO)4] 0 sp3 Diamagnetic Tetrahedral 4

28) (A)
Two bridging hydrogen atoms are perpendicular to the rest four H-atoms.
(B) B2H6 + 4MeCl B2Me4H2 + 4HCl
All terminal B–H bonds are broken and H are replaced by Me group.
(C) In B2H6 bridging H-atoms are in a plane perpendicular to the plane containing rest four
H-atoms (see option 'A' figure).
(D) All B–H bond distance are not equal as bridge bond length are longer than B-HT (Terminal)
bond lengths.
Thus, (D) option is incorrect.

29) is cationic compex where as is anionic complex.

30) Theory Based.

31) Theory Based.

32)

The correct answer option is (C)

33) pH = 5 ⇒ (H+) = 10–5M ; VL

pH = 3 ⇒ (H+) = 10–3M ; VL

finally [H+] =

=
pH = 3 + log 2 = 3.3
34) h =
% h = 10–4 × 100 = 0.01%

35) H2O H+ + OH–; ΔH = +ve (endothermic)

T ↑ ⇒ (H+) ↑ ⇒ pH ↓

36)
= 186.2 – 2 × 130.6 – 6
= – 81 J/K
ΔG° = ΔH° – TΔS° = –TΔSuni
–75000 – 300(–81) = –300 ΔSuni

37)

38) –2601 = 4 × ΔH°f [CO2] + 2 × ΔH°f [(H2O(ℓ)] – 2 × ΔH°f [C2H6]

–2601 = –394 × 4 – 285.8 × 2 – 2 × ΔH°f C2H6
= –1576 – 571.6 – 2 × ΔH°fC2H6
ΔH°f(C2H6) = 226.7 kJ/moles

39)

The correct answer is option (A)

40)

The correct answer option is (D)

41)

The correct answer option is (C)

42)

The correct answer option is (C)

43)

The correct answer option is (C)

44)

The correct answer option is (A)

45)

The correct answer option is (B)

46) Explanation - Naming of different ores.

Concept - Sulphide ores are concentrated by forth floatation method.
Solution -
Chromite ore : FeCr2O4 [Feo.Cr2O3] oxide ore
✓Galena : pbs [Sulphide ore]
✓Copper pyrites : CuFeS₂ [Sulphide ore]
Cassiterite : SnO₂ (Oxide)
Carnalite : KCl·MgCl₂·6H₂O [Halide]
✓Zinc blende : ZnS [Sulphide]
Dolomite : CaCo3.MgCo3 [Carbonate ore]
Final Answer - 3

47)

Number of electrons of Na for which |m| is 1

2 2 6 1
11Na = 1s 2s 2p 3s

m = –1 0 +1
|m| = 1 0 1
No. of electrons = 4

48) Let molarity of KMnO4 = x

KMnO4 + FeSO4 → Fe2(SO4)3 + Mn2+
n=5 n=1
(Equivalents of KMnO4 reacted) = (Equivalents of FeSO4 reacted)
⇒ 5 × x × 10 ml) = 1 × 0.1 × 10 ml
⇒ x = 0.02 M
Molar mass of KMnO4 = 158 gm/mol
⇒ Strength = (x × 158) = 3.16 g/ℓ

49) At equilibrium pressure in each chamber is the same.

n1 = = 10-2 ; n2 = = 5 × 10–2
Let A = area of cross section of cylinder.
 P= = ⇒ = =5 Ans.

50)

The correct answer is 4.

PART-3 : MATHEMATICS

51) Equation of normal to parabola

y = mx – 2m – m3
this normal passing through center of circle (2, 8)
8 = 2m – 2m – m3
m = –2
So point P on parabola ⇒ (am2, –2am) = (4, 4)
And C = (2, 8)
PC =
d2 = 20

52) A = |A| = –1
7 6 5
⇒ |A – (β – 1)A – βA | = 0
⇒ |A|5 |A2 – (β – 1)A – βI| = 0
⇒ |A|5 |(A2 – βA) + A – βI| = 0
⇒ |A|5 |A(A – βI) + I(A – βI)| = 0
|A|5 |(A + I) (A – βI)| = 0

A+I= ⇒ |A + I| = –4,
Here |A| ≠ 0 & |A + I| ≠ 0

A – βI =

|A – βI| = 2 – 3(1 – β) = 3β – 1 ⇒
9β = 3
53) A ≡ (4, 4) here y2 = 4x
(at12, 2at1)
a =1
(t12, 2t1) ⇒ t1 = 2
Let co-ordinates of P be
(at22, 2at2) ≡ (t22, 2t2)
But normal again meet the curve at P.

∴ t2 = –t1 – = –2 = –2 – 1 = –3
∴ P(9, – 6)
Now co-ordinates of B (4, – 4)
(t32, 2t3) ⇒ t3 = –2
Let co-ordinates at P′ be (t42, 2t4)
But normal again meet the curve at P′

∴ t4 = = –(–2) – =2+1=3
∴ P′(9, 6)
Hence PP′ = 12
Statement 2 is wrong

54)
I.F. =

Let tan–1x = z ∴

∴ y.ez =

∵ f(1) = 0
∴

55) α (θ) = 1 – 2sin2 θ cos2 θ

Also, β(θ) = – (sin4 θ cos4 θ + (1 + cos2 θ) (1 + sin2 θ))
= – (sin4 θ cos4 θ + 1 + cos2 θ + sin2 θ + sin2 cos2 θ)

= –(t2 + t + 2), t =

⇒ β (θ) ≥

56) By dividing both the sides by cos2y. We get

Put

⇒ which is linear in z
The integrating factor is
I.F. =
Hence, the solution is

∴ (6 tan y) x2 = x6 + 6a
⇒ 6x2 tan y = x6 + c, [c = 6a]

57)

xy(xdx + ydy) = xdy + ydx

on integrating

58)
59) (A) Degree = 1

(B)
I=0
(C) By applying L'Hopital we get

(D)

60)
F(x) is continuous and differentiable
F(1) = F(2)
one root of
in (1, 2)

61) Given that,

I1 =

and
Since, for 0 < x < 1
and for x > 1

and
⇒ I2 < I1 and I4 > I3

62) Apply king and add

63) s1 < 0 ([p]+1)2+ 4 – [p] –1– 2 – 8 < 0 {∵ [p +1]= [p] +1}

[p]2 + [p] – 6 < 0
([p] + 3) ([p] – 2) < 0 ⇒ [p] ∈ (–3, 2)
∴ p ∈ [–2, 2) a = –2 , b = 2
 ⇒ +

⇒ 0+2 ⇒ 2 =2

64)

65) f(x) = cosx – x + 1

f ′ (x) = –sinx – 1
f is decreasing ∀ x ∈ R
f(x) = 0
f(0) = 2, f(π) = –π
f is strictly decreasing in [0, π] and f(0).f(π) < 0
⇒ only one solution of f(x) = 0
S1 is correct and S2 is incorrect.

66)

ƒ(x) = x(x – 1)|x(x – 4)(x – 5)|

So ƒ(x) is not differentiable at x = 4 & 5.

67) r = 1 and = 1 ℓn(ep) = 1 ⇒ p = 1, q = 0

68)
∴ From above graph of f(x), f(x) is one-one and onto both.

69)

70)

71) We have AB =
Now tr(AB) = tr(C) ⇒ 3ax2 + b + 6cx = (x + 2)2 + 2x + 5x2 ∀ x ∈ R (Identity)
⇒ 3ax2 + 6cx + b = 6x2 + 6x + 4
Hence a = 2, c = 1, b = 4

Consider I = ; (Put x = 2t , so O, dx = 2 dt ; to make coefficient of x2 and

constant term same)

= =

I2 = (Put t = , dt = – dy); I2 =

I2 = = = – I2 ⇒ I2 = 0
Now I1 = = = =

Hence I = = =
⇒ p = 2, q = 27
∴ Value of (p + q) = 2 + 27 = 29 Ans.

72)

y" =

⇒ y'(0) = 1

slope = 1
⇒ x = 0 is point of inflection

73) There can be infinitely many parabolas through given points.

∴ 12 (π – 4A) = 12

74) Tangent to parabola y = mx + is also tangent to hyperbola

∴ c2 = a2m2 – b2 ⇒ = m2 – 3 ⇒ m = ±2

∴ Area (Δ) = ⇒ 2Δ = 9
75) Using L'Hospital Rule