17-08-2025

0999DJA161103250006 JA

PART-1 : PHYSICS

SECTION-I (i)

1) A stone is projected from ground at t = 0. At the time of projection horizontal and vertical
component of velocity are 10 m/s and 20 m/s respectively. Then time at which tangential and normal
acceleration magnitude will be equal (g = 10 m/s2) [neglect air friction] :-

(A)

(B)

(C) 3 sec
(D) 4 sec

2) A particle of mass m moves along the internal smooth surface of a vertical cylinder of radius R as
shown. The force which acts on the wall of the cylinder if initially the velocity v0 of the particle
makes an angle α with the horizotal (Assume that particle does not leave contact with the curved

surface of the cylinder) is :-

(A)

(B)

(C)

(D) mg

3) In the circuit shown the cell will deliver maximum power to the network if the value of ‘R’ is :-
(A)

(B)

(C)

(D)

4) Two light bulbs shown in the circuit have ratings A (24 V, 24 W) and B (24 V and 36W) as shown.

When the switch is closed :

(A) the intensity of light bulb A increases

(B) the intensity of light bulb A decreases
(C) the intensity of light bulb B remain unchange
(D) the intensity of light bulb B decreases

SECTION-I (ii)

1) A block of mass 25 kg rests on a horizontal floor (μ = 0.2). It is attached by a 5m long horizontal

rope to a peg fixed on floor. The block is pushed along the ground with an initial velocity of 10 m/s
so that it moves in a circle around the peg. Choose the correct statement(s) :-

(A) Magnitude of tangential acceleration of the block is 2 m/s2

(B) Speed of the block at time t = 1 sec is 8 m/s
(C) Time when tension in rope becomes zero is 5 sec
(D) Friction force acts towards centre of the circle

2) A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves
as a conical pendulum with the string making 60° with the vertical. Then :

(A)
its period of revolution is sec.
(B) the tension in the string is doubled the weight of the particle
(C) the velocity of the particle = m/s
(D) the centripetal acceleration of the particle is 9.8√3 m/s2.

3) A vehicle is moving on a circular road which is rough and banked also. Situation of vehicle is as
shown. Consider car is moving out of the plane on a banked road. Select the correct statement(s):

(A) Friction 'f' will be towards 'A'. If v > .

(B) Friction 'f' will be towards 'B'. If v > .
(C) Friction 'f' will be towards 'A' if v < .
(D) Friction 'f' will be towards 'B' if v < .

4) If the current in resistance of uniform cross section is doubled then, (assume temperature to be
approximately constant).

(A) the current density is doubled

(B) the conduction electron density is doubled
(C) the mean time between collision is constant
(D) the electron drift speed is doubled

5) The figure shows the graph of potential gradient for two potentiometers x and y. Choose the

correct statement(s).

(A) y will be preferred over x for measuring potential.

(B) x will be preferred over y for measuring potential.
(C) Sensitivity of y is less than sensitivity of x.
(D) Sensitivity of x is less than sensitivity of y.
6) For given circuit, choose the correct option(s) :

(A) Reading of voltmeter is 50 V

(B) Reading of voltmeter is 60 V
(C) Reading of ammeter is 3A
(D) Reading of ammeter is 4A

SECTION-I (iii)

Common Content for Question No. 1 to 2

In circuit shown all ammeters are ideal

1) If switch S is open, the ammeter (s) that read more than 10 A

(A) A1 & A2
(B) A3 & A4
(C) A5 & A6
(D) None of these

2) If switch S is closed, the ammeter that show decrease in reading, is :-

(A) A1
(B) A2
(C) A5
(D) A6

Common Content for Question No. 3 to 4

A hemispherical bowl of radius R = 0.1 m is rotating about its own axis (which is vertical) with an
angular velocity ω. A particle of mass 10–2 kg on the frictionless inner surface of the bowl is also
rotating with the same ω. The particle is at a height h from the bottom of the bowl.”
3) Obtain the relation between h and ω :-

(A) h = R + (g/ω2)
(B) h = R–(g/ω2)
(C) h = R–(2g/ω2)
(D) None of these

4) The normal force on the mass when h = R/2 is :-

(A) 2 N
(B) 0.5 N
(C) N
(D) 0.2 N

SECTION-III

1) Two heaters operating on 220 V, can individually supply a certain amount of heat in 10 min and
15 min separately. If both of them are operated simultaneously in parallel on 220 V, then in how
many minutes, will the same amount of heat will be delivered?

2)

Find the current shown by ammeter (ideal) in ampere.

3) A potentiometer wire of resistance per unit length (where x is distance from left
end of the potentiometer wire) is connected to four ideal batteries as shown in the figure. Resistance
of connecting wires is negligible. A cell of e.m.f. 1 volt is balanced against potential drop on
potentiometer wire. How many null point can be obtained on the wire?
4) A solid cylinder of radius 1m and mass 10kg is free to rotate about its axis. A torque of 5πNm is
applied to it for a period of 3sec and then it is suddenly reversed in direction without changing the
magnitude. The angle rotated by the cylinder in time interval 6 to 10 sec. is θ radian. If θ = nπ, fill n
in the OMR sheet.

5) A body of mass m = 0.1 kg is connected to a massless and inextensible thread of length ℓ = 1 m.

It rotates in a vertical circle such that the center of circle is 2 m above the floor. When passing
through the lowest position, the thread breaks and the body falls on the floor at a distance of 4 m
(horizontal) from the point of breakage. Determine the tension (in N) in thread just before it breaks.

6) Two particles A and B perform circular motion as shown in diagram. The locations of the particles

are shown at t = 0. Find radius of curvature (in m) of B as seen by A at (R = 1m)

SECTION-IV

1) A particle is moving in uniform circular motion in x-y plane, with radius 1m and centre as origin.
At t = 0, the particle is (R, 0) as shown. Time period of the path is 4 seconds. Match the quantities of

column-I with possible instant of time(s) of column-II.

Column-I Column-II

till t = 1
(A) Direction of average acceleration is (P)
sec.

till t = 3
(B) Direction of average velocity is (Q)
sec.

till t = 5
(C) (R)
Distance travelled by the particle is sec.

Magnitude of displacement of the particle till t = 7

(D) (S)
is sec.

till t = 9
(T)
sec.
2) Column-I shows electric circuits and Column-II shows equivalent resistance between A and B.

Column-I Column-II

(A) (P) R/2

(B) (Q) R/3

(C) (R) 2R/3

(D) (S) 2R

(T) None

PART-2 : CHEMISTRY

SECTION-I (i)

1) xMnO4– + yCN– + wH2O → xMnO2 + yCNO– + zOH–

in the balanced chemical reaction x : y : z is:

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

2) When a equimolar mixture of Cu2S and CuS is titrated with KMnO4 in acidic medium, the final
product contains Cu2+, SO2 and Mn2+. If molar masses of compounds Cu2S, CuS and KMnO4 are M1,
M2 and M3 respectively then incorrect statement is

(A)
Equivalent mass of Cu2S is

(B)
Equivalent mass of CuS is

(C)
Equivalent mass of KMnO4 is
(D) Cu2S and CuS both have same number of equivalents in mixture

3) Number of geometrical isomer(s) of square planar complex [RhCl(PPh3)(H2O)(CO)] is-

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

4) The geometry of [Ni(CO)4] and [Ni(CN)4]2– are

(A) both square planar

(B) both tetrahedral
(C) tetrahedral and square planar, respectively
(D) square planar and tetrahedral, respectively.

SECTION-I (ii)

1) Calcium silicate slag formed in extraction of iron

(A) prevents the reoxidation of molten iron.

(B) catalyses the combustion of carbon.
(C) reduces CO2 to CO at the bottom of the furnace.
(D) is used in cement industry.

2) The major role of fluorspar (CaF2) which is added in small quantities in the electrolytic reduction
of alumina dissolved in fused cryolite (Na3AlF6) is

(A) as a catalyst
(B) to make the fused mixture very conducting
(C) to lower the melting temperature of the mixture
(D) to decrease the rate of oxidation of carbon at the anode.
3) Amphoteric nature of aluminium is employed in which of the following process for extraction of
aluminium?

(A) Baeyer’s process

(B) Hall’s process
(C) Serpec’s process
(D) Dow’s process

4) Which of the following complex is diamagnetic as well as inner orbital complex?

(A) [Co(OH2)6]SO4
(B) K4[Fe(CN)5(O2)]
–4
(C) [Mn(NCS)6]
(D) [Co(NH3)6]Cl3

5) Octahedral complexes of copper (II) undergo structural distortion (Jahn-Teller). Which one of the
given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine;
H2N – CH2 – CH2 – NH2)

(A) [Cu(H2O)6]SO4
(B) [Cu(en)(H2O)4]SO4
(C) cis-[Cu(en)2Cl2]
(D) trans-[Cu(en)2Cl2]

6) Metallurgical process of zinc involves roasting of zinc sulphide followed by reduction. Metallic
zinc distills over as it is volatile and impurities like Cu, Pb and Fe gets condensed. The crude metal
obtained is called spelter, which may be purified by

(A) electrolysis process

(B) fractional distillation
(C) polling
(D) heating with iodine

SECTION-I (iii)

Common Content for Question No. 1 to 2

At high temperature carbon reacts with water to produce a mixture of carbon monoxide, CO and
hydrogen, H2. C + H2O CO + H2 CO is separated from H2 and then used to separate nickel
from cobalt by forming a volatile compound, nickel tetracarbony, Ni (CO)4. Ni + 4CO Ni(CO)4

1) How many moles of Ni(CO)4 could be obtained from the CO produced by the reaction of 75.0g of
carbon ? Assume 100% reaction and 100% recovery in both steps.

(A) 6.25
(B) 1.563
(C) 3.125
(D) 25.0

2) Formation of volatile Ni(CO)4 and its subsequent heating gives pure Ni. Process is called -

(A) Hall
(B) Dow
(C) Serpeck
(D) Mond

Common Content for Question No. 3 to 4

The complex [Cr(NH3)6]+3 is a inner orbital complex with CFSE equal to –1.2Δ0. The Mn+2 and Fe+2
form outer orbital complex with NH3 ligand then.

3) The correct statement for the complex [Cr(NH3)6]+3 is :-

(A) It is diamagnetic with configuration

(B) It is paramagnetic with configuration

(C) It is paramagnetic with magnetic moment of order 4

(D) The statement about complex is incorrect

4)

The CORRECT statement for complex ammine of Fe(II) and Fe(III) is :

(A) Both Fe(II) and Fe(III) form outer orbital complex with NH3
(B) Both form inner orbital complex with NH3
(C) Fe(II) form outer while Fe(III) form inner orbital complex with NH3
(D) All are incorrect statements regarding the complex

SECTION-III

1) 20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the
presence of excess of KI using starch as an indicator. The molarity of Cu2+ is found to be _____ × 10–2
M
[nearest integer]
Given : 2Cu2+ + 4I– → Cu2I2 + I2
I2 + 2S2O32– → 2I– + S4O62–

2) Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with
evolution of CO2 gas. The number of sodium bromide molecules involved per molecule of sodium
bromate in the balanced chemical equation is.

3) A mixture containing equal moles of FeO and Fe0.8O was titrated with 70 ml 0.3M KMnO4 in acidic

medium. Millimoles of Fe3+ in final solution is 'x'. The value of is.

4) Find number of Co–N linkage in, Pentaamminecobalt(III)–µ–amidodiamminetriaquacobalt(III)
chloride.

5) How many of the following compound can be optically active ?

(a) [Co(en)2ClBr] (b) [Co(en)(NH3)2Cl2]+
(c) [Cr(NH3)4Cl2]+ (d) [Co(NH3)3Cl3]
(e) [Co(en)3]3+ (f) [Ni(dmg)2]
2–
(g) [Pb(EDTA)] (h) [Cr(ox)2(NH3)2]–
(i) [Cr(gly)3]

6) Among the following, total number of planar molecule(s)/ions are :

[PtCl4]2–, [Ni(NH3)6]2+, BCl3, [NiCl4]2–, SF4, [Ni(CN)4]2–

SECTION-IV

1) Match the anionic species given in Column-I that are present in the ore(s) given in Column-II

Column-I Column-II

(A) Carbonate (P) Siderite

(B) Sulphide (Q) Malachite

(C) Hydroxide (R) Bauxite

(D) Oxide (S) Calamine

(T) Argentite

2) A sample of raw material contain NaNO3 and NaIO3. The NaIO3 can be used as a source of iodine,
produced in the following reactions :
IO3– + HSO3– —→ I– + SO42– ..... (1)
I– + IO3– —→ I2 + H2O ..... (2)
One litre of sample solution containing 396 g of NaIO3 is treated with stoichiometric quantity of
NaHSO3. Now a substantial amount of same solution is added to reaction mixture to bring about the
reaction (2) to completion (Molar mass : I = 127, Na = 23, S = 32, O = 16)

Column-I Column-II
–
(A) n-factor of IO3 in reaction (2) (P) 6
–
(B) Number of moles of HSO3 used in reaction (1) (Q) 1,2

(C) Moles of I2 produced (R) 2

–
(D) Equivalents of IO3 used in reaction (2) (S) 5

(T) 3

PART-3 : MATHEMATICS
 SECTION-I (i)

1) The value of sin10° + sin20° + sin30° +......+ sin360° is -

(A) 1
(B) 0
(C) –1
(D) none of these

2) If 15sin4α +10cos4α = 6, for some α ∈ R,

then the value of 27sec6α + 8cosec6α is equal to

(A) 350
(B) 500
(C) 400
(D) 250

3) The solution of is -

(A) nπ

(B)

(C)

(D)

4) The number of solutions of the equation

6cos5θ – 6cos4θ – 5cos3θ + 5 cos2θ + cosθ – 1 = 0
(where 0 < θ < 360º) is

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

SECTION-I (ii)

1) If , where C is arbitrary constant, then -

(A)
(B)
(C)
(D)
2) If , where ƒ(1) = 1 and a,b ∈ N,
then (x > 0 and C is constant of integration)

(A) a > b
(B) b > a

(C)

(D)

3)

The value of the integral is equal to - (where C is constant of integration)

(A) cot–1(e–x) + C
(B) log (ex – e–x) + C
(C) tan–1 (ex) + C
(D) tan–1 (e–x) + C

4) equals - (where C is constant of integration)

(A)

(B)

(C)

(D)

5)

If then

(A)

(B)

(C)

(D)

6) The maximum value of log20(3sinx – 4cosx + 15) is less than

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

SECTION-I (iii)

Common Content for Question No. 1 to 2

Consider,
On the basis of above information answer the following :

1) The value of A + C is

(A)

(B)
–

(C)

(D) –

2) If and , then number of solutions of the equation g(x) = sgn(sin–1x),

where sgn(.) denotes signum function.

(A) 0
(B) 1
(C) 2
(D) 3

Common Content for Question No. 3 to 4

Consider an equation 4sin4θ + 4cos4θ = 1 – 4x2 – 4x.
On the basis of above information, answer the following questions :

3) Number of values of θ ∈ [0,2π] satisfying the given equation for some real value of x is-

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

4) Number of real values of x satisfying the given equation for some real value of θ, is-

(A) 0
(B) 1
(C) 2
(D) infinite
 SECTION-III

1) If = ,
where C is constant of integration, then is equal to

2) If = where C is
constant of integration and α ∈ N, then α is equal to :

3) Let where f(0) = 0. If where a, b ∈ N then

evaluate |a – b|.

4) Let
then is

5) The range of value's of k for which the equation 2cos4x – sin4x + k = 0 has atleast one solution is
[λ, µ]. Find the value of (9µ + λ)

6) If 0 ≤ x ≤ 2π, then find the number of solutions of the equation sin 2x = cos 3x.

SECTION-IV

1)

Column-I Column-II

(A) If where g(0) = 12ℓn2, then (P) 1

g(–1) is equal to

(B) If then (Q) 3

km is equal to

(C) Let , (R) 6

then ℓ + m is equal to
 Rational
(D) Let , then [g2(1)] (S)
number
is equal to (where [.] denotes greatest integer
function & C is constant of integration)

Irrational
(T)
number

2) Match the values of expression in Column-I with the values in Column-II

Column-I Column-II

(A) then (P) 1

the value of A+B is

(B) (Q) 5
then the value of 3A+2B is

(C) then the (R) 6

value of 8(B+2A) is

(D) (S) 8
then
the value of 3B–2A is

(T) 7
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

Q. 1 2 3 4
A. C B A B

SECTION-I (ii)

Q. 5 6 7 8 9 10
A. A,B,C A,B,C,D A,D A,C,D B,C A,C

SECTION-I (iii)

Q. 11 12 13 14
A. D B B D

SECTION-III

Q. 15 16 17 18 19 20
A. 6 1 2 8 9 9

SECTION-IV

Q. 21 22
A. A->PRT,B->QS,C->P,D->PQRST A->R,B->P,C->S,D->P

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 23 24 25 26
A. D D C C

SECTION-I (ii)

Q. 27 28 29 30 31 32
A. A,D B,C A,B D A A,B

SECTION-I (iii)

Q. 33 34 35 36
A. B D B C

SECTION-III

Q. 37 38 39 40 41 42
A. 4 5 5 9 6 3
 SECTION-IV

Q. 43 44
A. A->PQS,B->T,C->QR,D->R A->S,B->P,C->Q,D->R

PART-3 : MATHEMATICS

SECTION-I (i)

Q. 45 46 47 48
A. B D D D

SECTION-I (ii)

Q. 49 50 51 52 53 54
A. A,C B,C A,C A,C A,D B,C,D

SECTION-I (iii)

Q. 55 56 57 58
A. C A C B

SECTION-III

Q. 59 60 61 62 63 64
A. 0 7 2 2 7 6

SECTION-IV

Q. 65 66
A. A->S,B->PS,C->RS,D->QS A->P,B->Q,C->Q,D->R
 SOLUTIONS

PART-1 : PHYSICS

1)
at any instant
at = g sin θ
an = g cos θ
∵ an = at
⇒ tan θ = 1

⇒ =±1
Solving we get
t = 1 and 3

2)

3) The cell will deliver maximum power to the network if Req = r

4) Potential difference across bulb A decreases on closing the switch

⇒ intensity decreases
Potential difference across bulb B increases on closing the switch
⇒ intensity increases
5)
(A) Tangential acceleration is the retardation produced by the friction
a = –f/m = – μmg/m
at = – 0.2 × 10 = – 2 m/s2

(B) = at = – 2

=
v – 10 = – 2t
v = 10 – 2t
(C) Tension in the rope will become zero when centripetal acceleration becomes zero
i.e. when speed becomes zero
v = 0 ⇒ 10 – 2t = 0 ⇒ t = 5 sec

6) As T sin 60° = ........ (1)

T cos 60° = mg ........ (2)

∴ tan 60° =
v= Also, r = ℓ sin 60°

Time period = = = 2π

= 2π
From equation (1)
T = 2mg
v=

∴ Centripetal acceleration aC = = = g tan 60°

7) F.B.D. in frame of car

block will move upward and friction acts upward.

8) As i changes (in same resistor) ; n remain constant
by i = JA ; J become double
by i = nAeVd ; Vd become double

9) Value of is more for 'X' so 'X' will be preferred over 'Y' for measuring
potential.

10) i = 3A
and V = iR = (1) (50) = 50 V

11) Use KCL

Before connection
A1 = A2 = 9.5 A, A4 = A5 = 5A, A3 = A6 = 2
After connection
A1 = 12.5 A, A2 = 2.5 A, A3 =10A,
A4 = 7A, A5 = 8A, A6 = 5A

12) Use KCL

Before connection
A1 = A2 = 9.5 A, A4 = A5 = 5A, A3 = A6 = 2
After connection
A1 = 12.5 A, A2 = 2.5 A, A3 =10A,
A4 = 7A, A5 = 8A, A6 = 5A

13) N cos θ = mg
N sin θ = mω2R sin θ
N = mω2R
mω2Rcos θ = mg

(R – h) =

h=R –

N= = 2 × 10–2 × 10 = 0.2 N

14)

N cos θ = mg
N sin θ = mω2R sin θ
N = mω2R
mω2Rcos θ = mg
(R – h) =

h=R –

N= = 2 × 10–2 × 10 = 0.2 N

15)

Time resistance
∴ time ∝ resistance

16)
Find equivalent

17) Null point will be achieved when ΔV = 1V

∴ Total of 2 null points will be there.

18)

5π = × 10 × 1α
⇒ α = π rad/s2

θ = × at2 = 4.5 π rad. ; ω = αt = 3π rad/s

α' = –π ⇒ comes to rest in 0 = 3π – πt ⇒ 3sec

θ' = 3π × 3 – × π × 32 = 4.5π

In rest 4 sec θ'' = × α × t2 = 8π rad.

19)

4=
v2 = 80

T = mg+
T=9

20)

At
VB/A = 3ωR
aB/A = Rω2

R.O.C. = = 9R = 9

21) Avg acceleration =

(A) must be in the direction upward vertical to downward vertical

Average velocity =
Direction of average velocity in the direction of displacement.
(B) Particle should be in the 3rd and 4th quadrant
(C) Distance moved from starting point to quarter half (t = 1) sec.

(D) Magnitude R will be when particle is at circumference point of circle on y-axis.

22) (Q) Equivalent diagram

 (S)
Assigning potential to each corner & solving

PART-2 : CHEMISTRY

23) Balanced chemical reaction is

2MnO4– + 3CN– + H2O → 2MnO2
+ 3CNO– + 2OH–

24) Cu2S —→ Cu2+ + SO2

nF = 8
CuS —→ Cu2+ + SO2
nF = 6

25)

[RhCl(pph3) (H2O) (CO)] ⇒ [MABCD] type sq. planar ⇒ 3G.I

26) Ni → d10 → sp3

Ni+2 → d8 → dsp2

27) The impurities are removed react with calcium oxide to make a liquid slag that floats on
top of the molten iron and prevents the reoxidation of molten iron. The slag is collected after
the denser iron has been run out of a top hole near the bottom of the furnace.

28) In the electrolytic reduction of Al2O3 cryolite (Na3AlF6) is added with CaF2 (fluorspar) to :-
• decrease m.p. of Al2O3
• decrease viscocity of electrolyte (CaF2 is used)
• increase conductivity
Hence option B,C are correct option.

29) (A) Baeyer's process : REACTION WITH NaOH (Acidic nature of Al reflects)
(B) Hall's process : REACTION WITH Na2CO3 (Acidic nature of Al reflects)
(C) Serpeck' process : REACTION WITH (N2 + C) (Acid & Base are not used in this reaction )
(D) Dow's process : NOT USED FOR ALUMINIUM EXTRACTION

30)

Concept : Inner orbital complex = d2sp3 hybrid diamagnetic ⇒ No unpaired e–

(A) [Co(H2O)6]2+ Co2+ 3d7 has atleast on unpaired e– (paramagnetic)
(B) [Fe (CN)5(O2)]4– O2– : Super oxide ligand, has unpaired e– paramagnetic
(C) [Mn(NCS)6]4– Mn2+ 3d5 has atleast on unpaired e– (Paramagnetic)
(D) [Co (NH3)6]3+ ⇒ Co3+ : 3d6
NH3 : SFL (Pairing of e–)
t2g222 eg00
d2sp3 hybrid, inner orbital complex, Diamagnetic

31) Chelation (ring formation tendency) Jahn – Teller

So, maximum structural distortion will be maximum in [Cu(H2O)6]SO4

In above complex there is no chelating ligand.

32) Zinc spelter contains Pb, Fe, Cd, As, etc as impurities. Impure zinc can be purified by
following methods.
• By fractional distillation :-
The b. pt. of Pb, Fe are higher than that of zinc while that of cadmium, arsenic are lower than
that of zinc. When distillation is carried out around 1000°C, zinc, Cd, As, etc. distill off leaving
Pb and Fe the distillate is then heated to 800°C where Cd and As distill off leaving pure zinc.
This sample of Zn is about 99% pure.
• By electrolysis :-
Zinc of higher purity can be obtained by electrolysis. Pure zinc rod is used as cathode while a
block of impure zinc is used as anode. A mixture of ZnSO4 and dill H2SO4 is used as electrolyte.
On passing current impure zinc dissolves and equivalent amount of pure zinc is deposited at
cathode.

33) C + H2O CO + H2

Ni + 4CO Ni(CO)4

Mole
Moles of Ni(CO)4 = × moles of CO = 1.563 Mole

34) The Mond process, sometimes known as the carbonyl process, is a technique to extract
and purify nickel. This process involves the fact that carbon monoxide combines with nickel
radily and reversibly to give nickel carbonyl.

35)

36)

37) neq. of I2 = neq of Na2S2O3 = 20 × 0.002 × 1

2 × nmol of I2 = 0.4
nmol of I2 = 0.2 m mol
nmol of Cu+2 = 0.2 × 2 × 10–3

[Cu+2] =

38) It is type of disproportionation reaction

(i) Br2 → Br– +
(ii) Br2 + Br2 → Br– +

(iii)
3Br2 + 3Na2CO3 → 5NaBr + NaBrO3 + 3CO2

39) let mmoles of each is = x

n-factor of FeO = 1
n-factor of Fe0.80O = 0.4
meq of FeO + meq of Fe0.80O = eq of KMnO4
x × 1 + x × 0.4 = 70 × 0.3 × 5
x = 75 mmoles
mmoles of Fe3+ produced
= 75 + 75 × 0.8 = 135 mmoles

40)

[(NH3)5 Co NH2Co(NH3)2 (H2O)3] Cl5

Total Co - N linkage are = 9

41) a,b,e,g,h,i
(a) Cis-isomer is op. active

(b) is op. active

(c) optically inactive.
(d) op. inactive
(e) always optically active
(f) square planar, optically inactive
(g) EDTA complexes are optically active
(h) cis - active
(i) Both geometrical isomerism are optically active.

42) [PtCl4]2–, BCl3, [Ni(CN)4]2– are planar.

43) (A) → P, Q, S,
(B) → T,
(C) → Q, R,
(D) → R
Siderite : FeCO3
Malachite : CuCO3.Cu(OH)2
Bauxite : Al2O3.2H2O consisting of part of hydroxide of aluminium also and the general formula
is
AlOx(OH)3–2x
where 0 < x < 1
Calamine : ZnCO3
Argentite : Ag2S
 44) (Q) gm eq. of IO3– = gm eq. of HSO3– = gm eq. of I–

× 6 = moles of HSO3– × 2 = moles of I– × 6

(R) gm eq. of I2 = gm. eq. of I–

moles × = 2 × 1
moles of I2 = 1.2

PART-3 : MATHEMATICS

45) sin 10º + sin 20º + sin 30º + ... + sin 360º
= sin 10º + sin 350º + sin 20º + sin 340º + ...... (sin 180º + sin 360º)
= (sin 10º – sin 10º) + (sin 20º – sin 20º) + ....... (0 + 0)
=0

46) 15sin4α + 10cos4α = 6

15sin4α + 10cos4α = 6(sin2α + cos2α)2
(3sin2α – 2cos2α)2 = 0

⇒ 27sec6α + 8cosec6α
= 27(sec6α)3 + 8(cosec6α)3
= 27(1 + tan2α)3 + 8(1 + cot2α)3
= 250

47) LHS = (on simplifying)

& RHS =

⇒
⇒ θ =nπ
But

48) 6cos4θ(cosθ – 1) – 5cos2θ(cosθ – 1) + (cosθ – 1) = 0

(cosθ – 1) (6cos4θ – 5cos2θ + 1) = 0

Number of solution = 8

49)

put sinx = t
⇒ F(x) =

∵ ⇒C=2

⇒ F(x) = 2 –

50)

put

⇒ ƒ(x) = x +

51)
Let

52)

let
Option AC is correct.

53)

sin α + sin β = a .....(i)

cos α – cos β = b ......(ii)
(i)2 + (ii)2
2 – 2 cos (α + β) = a2 + b2

sin α + sin β = a

............. (i)
cos α – cosβ = b

..............(ii)
(ii) + (i)

54) log20 (3 sin x – 4 cos x + 15)

–5 ≤ 3 sin x – 4 cos x ≤ 5
– 5 + 15 ≤ 3 sin x – 4 cos x + 15 ≤ 5 + 15
10 ≤ 3 sin x – 4 cos x + 15 ≤ 20
So max value is

55)

= =

= =

A+C=

56) ƒ (x) = –2x

g(x) = –x2 + c'

∴ g(x) = – x2 +
sgn (sin–1x) = –1 sin–1 x < 0 ∀ x ∈ [–1, 0)
Both curves do not intersect in
∴ no solution is there.

57)

4(sin4θ + cos4θ) = 1 – 4x2 – 4x

⇒ 4(1 – 2sin2θcos2θ)

Now
and 2 – (2x + 1)2 ∈ (–∞, 2]

or x = , sin22θ = 1

or 2θ = nπ ±

or θ =

58)

4(sin4θ + cos4θ) = 1 – 4x2 – 4x

⇒ 4(1 – 2sin2θcos2θ)

Now
and 2 – (2x + 1)2 ∈ (–∞, 2]

or x = , sin22θ = 1

or 2θ = nπ ±

or θ =

59)
=
∴ α = – 9, β = 3
⇒ (α + 3β) = 0

60)

Put cosecx + cotx = t ⇒ cosecx – cotx =

⇒ (–cosecx cotx – cosec2x)dx = dt

α=7

61)

Let

then

Where
∵ f(0) = 0, ∴ c = 0

∴ |a – b| = |10 – 12| = 2

62)

(1 + tanθ) (1 + tan(45º – θ)) = 2

((1 + tan1º) (1 + tan4º)) ((1 + tan2º) (1 + tan4º))
.... (1 + tan22º) (1 + tan23º)
= 222 ⇒ λ = 22

63) sin4x – 4sin2x + (2 + k) = 0

Let sin2x = t t ∈ [0, 1]
2
t – 4t + (2 + k) = 0
f(0) f(1) ≤ 0
(k + 2) (k – 1) ≤ 0 ⇒ –2 ≤ k ≤ 1

64)

sin2x = cos3x

n = 0, ,

65) (A)
Let x = t3
dx = 3t2dt

I=

g(x) =

g(–1) =

(B) I=
 =
Let tan x = t5
sec2x dx = 5 t4 dt

⇒ I= = =
⇒ km = 1

(C)

⇒
⇒ ℓ+m=6

(D)
Let 1 + xex = t2
⇒ (x + 1)exdx = 2tdt

⇒
⇒ g2(x) = 1 + xex ⇒ g2(1) = 1 + e
⇒ g2(1) = 3.71828 ⇒ [g2(1)] = 3

66) (A)

then A = A + B =1
(B) Let 3sin x + 4cos x = A(2 sin x + cos x) + B(2cos x – sin x)
⇒ 2A – B = 3 and A + 2B = 4
solving A = 2, B = 1
(C) Let 3ex – 5e–x = A(4ex + 5e–x) + B(4ex – 5e–x)
Compairing 4(A+B) = 3
5(A– B) = –5

⇒ A= , 8(B + 2A) = 5

(D) let tan x = t

A = –2, B = , 3B – 2A = 6