11-12-2024

9610WJA801005240118 JA

PART 1 : PHYSICS

SECTION-I

1) A ball of mass 5kg moving with speed 8 m/s collides head on with another stationary ball of mass
15 kg. if collision is perfectly inelastic, then loss in kinetic energy is

(A) 160 J
(B) 80 J
(C) 40 J
(D) 120 J

2) where
λ = Wavelength
n1 and n2 = orbit number then Rydberg's constant R is :-

0
(A) M L–1T
(B) MLT–1
0 0
(C) M L–1T
0
(D) ML T2

3) An ideal gas is expanding such that PT2 = constant. the coefficient of volume expansion of the gas
is :

(A)

(B)

(C)

(D)

4) A cylindrical tube of cross-sectional area A has two air tight frictionless pistons at its two ends.
The pistons are tied with a straight piece of metallic wire. The tube contains a gas at atmospheric
pressure P0 and temperature T0. If temperature of the gas is doubled then the tension in the wire is :-

(A) 4 P0 A
(B) P0A/2
(C) P0 A
(D) 2 P0 A

5) Water level is maintained in a cylindrical vessel upto fixed height H, as shown. At what height
above bottom a hole should be made so that water stream coming out strike bottom at maximum

distance from vessel :-

(A)

(B)

(C)

(D) H

6) A spaceship is moving with constant speed v0 in gravity free space along +Y‑axis suddenly shoots
out one third of its part with speed 2v0 along + X‑axis. Find the speed of the remaining part.

(A) v0

(B) 0
v

(C) 0
v

(D) 0
v

7) Consider three vectors , , and . A vector of the form

(α and β are numbers) is perpendicular to . The ratio of α and β is

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

8) In the arrangement, spring constant k has value 2Nm–1, mass M = 3 kg and mass m = 1 kg. Mass
M is in contact with a smooth surface. The coefficient of friction between two blocks is 0.1 and
amplitude of oscillation is 10 cm. The time period of SHM executed by the system is :-
(A)
(B)
(C)
(D) 2π

9) A 72 Ω galvanometer is shunted by a resistance of 8 Ω. The percentage of the total current which

passes through the galvanometer is :

(A) 0.1%
(B) 10 %
(C) 25%
(D) 0.25%

10) A concealed circuit (a black box) consisting of resistors has four terminals. If a volatage is
applied between 1 and 2 when clamps 3 and 4 are open, the power liberated is P1 = 40 W and when
clamps 3 and 4 are connected the power liberated is P2 = 80 W. If the same source is connected to
the clamps 3 and 4, the power liberated in the circuit when clamps 1 and 2 are open is P3 = 20 W.

Determine the power P4 consumed in the circuit when

the clamps 1 and 2 are connected and the same voltage is applied between the clamps 3 and 4.

(A) 20 W
(B) 40 W
(C) 60 W
(D) 80 W

11) A large sheet carries uniform surface charge density σ. A rod of length 2l has a linear charge
density λ on one half and -λ on the second half. The rod is hinged at mid-point O and makes angle θ

with the normal to the sheet. The electric force experienced by the rod is

(A) 0

(B)

(C)
(D) None of these

12) A positive point charge Q is kept (as shown in the figure) inside a neutral conducting shell whose

centre is at C. An external uniform electric field E is applied. Then

(A) force on Q due to E is zero

(B) net force on Q is zero
(C) net force acting on Q and conducting shell considered as a system is zero
(D) net force acting on the shell due to E is zero

13) A conducting sphere (of radius R) has concentric spherical cavity (radius ) with a charge (–q) at
its centre. A charge (+q) is placed at a distance 2R from the center. Mark the incorrect option :-

(A) Charge density on the outer surface is not uniform

(B) Charge density on the inner surface is uniform.

(C)
Electric field at the centre due to point charge (+q) is in magnitude.
(D) Potential difference between inner and outer surface of sphere is zero.

14) A large number of identical point masses m are placed along x-axis at x = 0, 1, 2, 4, ....... The

magnitude of gravitational force on mass at origin (x = 0), will be :-

(A) Gm2

(B)
Gm2

(C)
Gm2

(D)
Gm2

15)

A satellite is in a circular orbit very close to the surface of a planet. At some potential it is given an
impulse along its direction of motion, causing its velocity to increase n times. It now goes into an
elliptical orbit. The maximum possible value of n for this to occur is

(A) 2
(B)
(C) +1

(D)

16) A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel
and 15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The
copper end of the composite rod is maintained at 100°C and the aluminium end at 0°C . The whole
rod is covered with belt so that there is no heat loss occurs at the sides. If KCu = 2KAl and KAI = 3KNi ,
then what will be the temperatures of Cu–Ni and Ni–Al junctions respectively :-

(A) 23.33°C and 78.8°C

(B) 83.33° and 20°C
(C) 50°C and 30°C
(D) 30°C and 50°C

17) Two spheres each of mass M and radius R/2 are connected with a massless rod as shown in the
figure. What will be the moment of inertia of the system about an axis passing through the centre of

one of the spheres and perpendicular to the rod.

(A)

(B)

(C)

(D)

18) A cylinder of radius 10 cm rides between two horizontal bars moving in opposite directions as
shown in the figure. The location of the instantaneous axis of rotation and the angular velocity of the

roller are, respectively, (There is no slipping at P or Q)

(A) 15 cm from Q, 2.50 rad/s

(B) 8 cm from Q, 1.25 rad/s
(C) 8 cm from Q, 2.50 rad/s
(D) 15 cm from Q, 1.25 rad/s

19) A stone is thrown upward with a speed 'u' from the top of the tower & it reaches the ground with
speed '3u'. The height of the tower is :-

(A)

(B)

(C)

(D)

20) Water flows through the tube shown. Area of cross-section of wide and narrow part are 5 cm2 &
2 cm2. The rate of flow is 500 cm3/sec. Find difference in mercury level of U-tube :-

(A) 2.9 cm
(B) 1.9 cm
(C) 0.9 cm
(D) None of these

SECTION-II

1) A thin insulating uniformly charged (linearly charged density λ) rod is hinged about one of its
ends. It can rotate in vertical plane. If rod is in equilibrium by applying vertical electric field E as
shown in figure. Find the value of E(in N/C). (Given that mass of rod 2 kg, , l =1m, g=10

m/s2)

2) A non-conducting uniformly charged spherical shell is kept on a rough non-conducting horizontal

surface. Mass of shell is 2kg and charge on it is 4C and friction coefficient between surface and shell
is 0.18. Take g = 10m/s2. The minimum value of external uniform horizontal electric field so that

shell starts rolling on the surface is E0 in SI units. If E0 is N/C. Then the value of x is.
3)

The gravitational field in a region is given by N/kg. Find out the work done (in joule) in
displacing a particle by 1 m, along the line 4y = 3x + 9

4) In order to measure the internal resistance r1 of a cell of emf E, a meter bridge of wire resistance
R0 = 50 Ω, a resistance R0/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are
used in a circuit, as shown in the figure. If the null point is found at =72 cm, then the value of r1 =

____ Ω.

5) A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω

resistance, it can be converted into a voltmeter of range 0 – 30 V. If connected to a resistance,

it becomes an ammeter of range 0 – 1.5 A. The value of n is–

PART 2 : CHEMISTRY

SECTION-I

1) From 3.01 × 1024 molecules of water, 0.5 g-molecules of water are removed, then volume of water
remain is :–

(A) 89.5 ml
(B) 81 ml
(C) 35.5 ml
(D) 30 ml

2) The ionisation energy of a certain element is 1540 kJ mol–1. When the atoms of this element are in
the first excited state, the ionisation energy is only 300 kJ mol–1. The region of the electromagnetic
wave spectrum in which the light is emitted in a transition from the first excited state to the ground
state.

(A) Visible
(B) UV
(C) IR
(D) X-Ray

3) The ratio of time periods in first and third orbits of hydrogen atom is :-

(A) 1 : 3
(B) 1 : 4
(C) 1 : 27
(D) 1 : 9

4) The order of basic character of given oxides is :

(A) Na2O > MgO > Al2O3 > CuO

(B) MgO > Al2O3 > CuO > Na2O
(C) Al2O3 > MgO > CuO > Na2O
(D) CuO > Na2O > MgO > Al2O3

5) Which of the following is the correct order of ionisation enthalpy ?

(A) Te2– < I– < Cs+ < Ba2+

(B) I– < Te2– < Cs+ < Ba2+
(C) Te2– < Cs+ < I– < Ba2+
(D) Ba2+ < Cs+ < I– < Te2–

6) The pair in which both species have same hybridisation and same molecular shape both :

(A) CH4 , H2O

(B) PCl5 , SF4
(C) XeF4 , ICl2¯
+
(D) PCl4 , SiF4

7) For the equilibrium,

2H2O H3O+ + OH–, the value of ΔG° at 298 K is approximately :-

(A) –80 kJ mol–1

(B) –100 kJ mol–1
(C) 100 kJ mol–1
(D) 80 kJ mol–1

8) 0.25 mole of solid "A" is taken in a 1 litre evacuated flask at 27°C. Following two equilibrium exist
simultaneously.
Let 0.2 moles of solid decomposes upto equilibrium & 0.1 mole of
E was found at equilibrium.
[Given : R = 0.08 L atm K–1 mol–1]
Calculate partial pressure of B at equilibrium.

(A) 1.2 atm

(B) 0.8 atm
(C) 2.4 atm
(D) None of these

9) In the system if the concentration of Ca2+ is increased by four

times, then the equilibrium concentration of will be changed to :

(A) One half of its initial value

(B) Twice the initial value
(C) One fourth of its initial value
(D) Thrice of its initial value

10) A certain weak acid has Ka = 1 × 10–4. Calculate the equilibrium constant for its reaction with
strong base -

(A) 106
(B) 108
(C) 1010
(D) 10–10

11)

Find pH of the resultant solution formed by the addition of 500 ml 0.1 M Ba(OH)2 and 500 ml 0.6 M
NH4Cl in pure H2O at 25°C (Use log2 = 0.3, log3 = 0.5, log5 = 0.7, Kb(NH3) = 10–5 ).

(A) 8.7
(B) 9.3
(C) 12
(D) 12.7

12) An oxidising agent react completely with 100 ml of acidified 0.1M KI (aq.) then for which of the
following oxidising agent, moles used will be least -

(A) KMnO4
(B) K2Cr2O7
(C) MnO2
(D) KIO3

13) Which of the following reactions are disproportionation reactions?

(A) Cu+ → Cu2+ + Cu
(B)
(C) 2KMnO4 → K2MnO4 + MnO2 + O2
(D)
Choose the correct answer from the options given below:

(A) (A), (B)

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

14) The reduction potential of hydrogen half-cell will be negative if :

 ⊕
(A) p(H2) = 2 atm [H ] = 1.0 M
⊕
(B) p(H2) = 2 atm and [H ] = 2.0 M
⊕
(C) p(H2) = 1 atm and [H ] = 2.0 M
⊕
(D) p(H2) = 1 atm and [H ] = 1.0 M

15) Equivalent conductance of saturated solution of BaSO4 is 400 ohm–1cm2eq–1 and specific
conductance is 8 × 10–5 ohm–1cm–1. Ksp of BaSO4 is :

(A) 4 × 10–8
(B) 1 × 10–8
(C) 2 × 10–4
(D) 1 × 10–4

16)

For a zero order reaction

A → B ; K = 0.025 Ms–1
If 2M of A is taken initially the concentration of A after 40 sec is -

(A) 0.5M
(B) 1 M
(C) 0.25 M
(D) 2M

17) 2Na2CrO4 + H2SO4 → Na2SO4 + Na2Cr2O7 + H2O.

In above reaction which of the following compound crystallise out first -

(A) Na2Cr2O7
H2O
(B)

(C) Na2SO4
(D) Na2Cr2O7 and Na2SO4 equally

18)

Compare C–O & M–C bond length in the following complexes.

(a) [M(CO)x]2+ (b) [M(CO)x]+
0
(c) [M(CO)x] (d) [M(CO)x]–

(e) [M(CO)x]2–
(A) for C–O a < b < c < d < e ; for M–C a > b > c > d > e
(B) for C–O a > b > c > d > e ; for M–C a > b > c > d > e
(C) for C–O a > b > c < d < e ; for M–C a < b < c > d > e
(D) for C–O a > b > c > d > e ; for M–C a < b < c < d < e

19) A hydrated colorless solid (A) is water soluble and finds use in medicine as a purgative. When a
solution of (A) is treated with ammonium phosphate, a white precipitate is formed. (A) gives a pink
mass in cobalt nitrate cavity test. What is (A)?

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

20) Total number of functional groups present in above

compound

(A) 7
(B) 8
(C) 9
(D) 6

SECTION-II

1) For the reversible heterogeneous reaction,

2A(s) B(s) + 2C(g)
the following values are given at 727°C
(ΔHf°) standard (Sm°) standard
enthalpy of molar.
Formation(kJ/mol) Entropy(J/K)
A –150 30
B –230 70
C –180 120
Then find Value of lnKp for the reaction at 727°C. (Given R = 8 J mol–1 K–1)
Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

2) The number of moles of KMnO4 required to reduce 1 mole ferrous oxalate in acidic medium is .
The value of x is:

3) A first order reaction has the rate constant, k = 4.6 × 10–3 s–1. The number of correct statement/s
from the following is/are ________.
Given : log 3 = 0.48
A. Reaction completes in 1000 s.
B. The reaction has a half-life of 500 s.
C. The time required for 10% completion is 25 times the time required for 90% completion.
D. The degree of dissociation is equal to (1 – e–kt).

4) Total number of carbonyl ligands which are present as bridging ligand in Fe2(CO)9 and Mn2(CO)10
are x and y respectively, then value of x + y ?

5) How many statements is/are correct ?

(i) Ag2SO3 is insoluble in dilute HNO3 .
(ii) Nitrogen dioxide (NO2) is absorbed by ferrous sulphate solution forming brown solution.
(iii) Heavy metal chlorides like AgCl, HgCl2. etc. do not respond to chromyl chloride test.
(iv) The group reagent for III group basic radicals is NaOH.
(iv) Green colour of alkaline tetrahydroxidochromate(III) becomes yellow on adding H2O2.

PART 3 : MATHEMATICS

SECTION-I

1) is

(A)

(B) cot–1(3)
(C) tan–1(3)

(D)

2)

Number of roots of the equation is

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

3) Product of all solutions of the equation is

(A) – 2
(B) – 4
(C) 8
(D) 16

4) Number of integers in the domain of the function

(where [.] and {.} denote greatest integer function and fractional part function respectively)

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

5)

0
0

If , then the value of is (where ncr = 0 ∀ r > n and c = 0) :-

(A) 1
(B) 2
(C) 4
(D) 2015

6) The coefficient of x100 in the expansion of is

(A) 1
(B) 200
100
(C) C2
(D) (2)100

7) equals to :

(A) (m+1)!
m+1
(B) Cm
(C) m!
m
(D) Pm–2

8) If (3x3 – 4x2 + 5x – 3)15 = a0 + a1x + a2x2 + ............. + a45x45. Find the value of a1 + 2a2 + 3a3 +
..... + 45a45.

(A) 1
(B) 150
(C) 15
(D) 90

9) The total number of numbers of the form a1a2a3a4a5 where a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 or a1 ≥ a2 ≥ a3 ≥

a4 ≥ a5 and a1, a2, a3, a4, a5 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9} is
(A) 2574
(B) 2565
(C) 1287
(D) 1278

10) 3 women and 15 men are to be arranged in a row such that there should be atleast 2 men
between all the two consecutive women. Then number of such arrangements is :

14
(A) C4 3!
(B)
14
(C) C12
(D) 14C3

11) The ratio of number of rectangles (not square) and number of squares in a chess board is.

(A) 2 : 1
(B) 1 : 2

(C)

(D) None of these

12) The number of ways in which the letters of the word 'TWITTER' can be arrange in the adjacent

figure such that no horizontal row is empty (each box contains at most one letter), is

(A)
(10C3 – 9C2).

(B)

(C)

(D)

13)

In a set of real numbers a relation R is defined as x R y such that then relation R is

(A) reflexive and symmetric but not transitive
(B) symmetric but not transitive and reflexive
(C) transitive but not symmetric and reflexive
(D) none of reflexive, symmetric and transitive

14) Let R1 be a relation on set of real numbers R such that

R1 = {(a, b), a, b ∈ R : be a irrational number}, then

(A) R1 is reflexive, symmetric but not transitive.

(B) R1 is reflexive, transitive but not symmetric.
(C) R1 is neither symmetric nor transitive.
(D) R1 is equivalence relation

15) Let n be a fixed positive integer. Define a relation R in the set Z of integers by aRb if and only if
a – b is divided by n. The relation R is

(A) reflexive
(B) symmetric
(C) transitive
(D) an equivalence relation

16) Let N denotes the set of all natural numbers and R be the relation on N × N defined by :
(a, b) R (c, d) ⇒ ad(b+ c) = bc (a + d), then R is :

(A) Reflexive, Symmetric but not Transitive

(B) Symmetric, Transitive but not Reflexive
(C) Transitive, Reflexive but not Symmetric
(D) Equivalence

17) In ΔABC if , c = 1 and A = 30° then nature of triangle is

(A) right angled

(B) right angled isosceles
(C) Isosceles
(D) scalene

18) In ΔABC, if 27Δ2 = s4 and R = 4 then area of ΔABC (symbols used have usual notation in triangle
ABC)

(A)
(B)
(C) 24
(D) 48
19) Let Sn = 1⋅(n –1)+2⋅(n –2)+3⋅(n –3)+… +(n –1) ⋅ 1, n ≥ 4. The sum is equal
to:

(A)

(B)

(C)

(D)

20) The sum of the series is equal to

(A)

(B)

(C)

(D)

SECTION-II

1) The number of real solutions of + sin–1 is :-

2) In a shooting competition a man can score 5, 4, 3, 2 or 0 points for each shot. Then the number

of different ways in which he can score 30 in seven shots is , the value of is

3) Let and f = P – [P], where [P] denotes the greatest integer function,then the value

of – 720 is

4)

If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)2022 is divided by 8 is
_______.

5) If the angle A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the
lengths of the sides opposite to A, B and C respectively, then the value of expression
is
 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. D C C C A B A C B B A D C B B B A B B B

SECTION-II

Q. 21 22 23 24 25
A. 2 9 0 3 5

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

SECTION-II

Q. 46 47 48 49 50
A. 4 3 2 3 2

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

SECTION-II

Q. 71 72 73 74 75
A. 2 7 2 1 3
 SOLUTIONS

PART 1 : PHYSICS

1)

2)

3) Given process equation is, PT2 = constant

Using PV = nRT

we can have, = constant

or = constant
By differentiating above equation on both sides,

or
In above equation left side term is representing coefficient of volume expansion of gas, as from
the definition. The coefficient of volume expansion is defined as fractional change in volume
per unit rise in temperature.

4)

Pv = nRT
In this case v is constant
therefore it T is doubled
then P also doubles
Tension = Pressure × Area
Net pressure = gas pressure
– atmospheric pressure
Pnet = 2P – P = P
new tension = PA

5)

Let hole is at highth

range

for maxima
 H – 2h = 0

h=

6)
Gravity free
Fext = 0
⇒ P ⇒ conserved
Along x axis
⇒ 0 = m . 2v0 + 2mv1
⇒ v1 = –v0
Along y axis ⇒ 3mv0 = 2mv2

⇒ v2 =

7)

8) Both blocks will perform SHM as µg ≥ ω2A

9)

10) When 1 and 2 are connected to V and 3 and 4 are open.

When 3 and 4 are closed and 1 and 2 connected to V,

So,

So,

When 3 and 4 are connected to V and 1 and 2 are open,

When 3 and 4 connected to V and 1 and 2 closed

=
So,

Dividing, and
Hence,

11) Nearby the plate, field is uniform. Equal and opposite forces are experienced by upper half
and lower half portions of the rod.

12) The net charge on the conducting shell is zero. The force on charge Q of any size placed in
an uniform electric field E is QE. Hence the force on the conducting shell due to external
electric field is zero.

13)

Properties of conductor

14) Fnet = Gmm

15)

nmax=

16) If suppose and

Since all metal bars are connected in series

So

and
⇒
Hence, if

⇒
⇒

Similar if

⇒
⇒

17)

18)

19) v2 = u2 + 2as
(–3u)2 = u2 + 2 (–g) (-h)

h=
20)

P1 – P2 = ρ

h × 13600 × g = × 1000

A1V1 = A2V2 =

21)
mg = qE

22)
F – f = ma

f .R = ⋅α

= ⋅

⇒ f= ma

∴ F = qE = f + ma = f≤ μmg

∴ E≤ . = 2.25
23)

For the line 4y = 3 x + 9

4dy = 3dx; 4dy – 3 dx = 0....(i)
For work in the region,

dW = =
= 3dx–4dy (from equation (i)) = 0

24) ....(1)
So R1 = 36Ω; R2 = 14Ω

Current I = ....(2)
Applying KVL in right side loop

r1 = 3Ω Ans. (3)

25)

For voltmeter
Ig = 0.006 A
V = [G + R]Ig

–R=
For ammeter
GIg = R(I – Ig)

; x=5
PART 2 : CHEMISTRY

26)

0.5 gm - molecule = 0.5 mole of water removed of water removed

∴ No. of molecule = No. of mole × NA
3.01 × 1024 = No. of mole × 6.02 × 1023
[∴ NA = 6.02 × 1023]
No. of mole = 5
Mole of water remaining = Total Mole - Mole removed
= 5 - 0.5
= 4.5 mole
Man of water remaining = mole × Mw(H2O)
= 4.5 × 18 = 81 gm

∴ density of water =
So volume of water remaining = 81 ml

27)

28)
Z → Constant

29) Basic character ∝ Metallic character.

30)

All are isoelectronic species but as number of protons i.e. atomic number increases, the
attraction between electron (to be removed) and nucleus increases and thus ionisation
enthalpies increase.
Order of Z : Te2– (52) < I– (53) < Cs+ (55) < Ba2+ (56). So same will be the order of IE.
31) PCl4⊕ → sp3 hybridisation , tetrahedral shape
SiF4 → sp3 hybridisation , tetrahedral shape

32) = –RT lnk = × 298 × 2.303 log10–14 = 79.9 KJ/moL

33) ...(i)
At eq. 0.05 0.1 0.2
...(ii)
0.1 0.05 0.1

34) The expression for the equilibrium constant is K = .

Let X and Y be the initial concentrations of ions.
The expression for the equilibrium constant will be K = = XY2 ........ (1)
When the concentration of the Ca2+ ions is increased four times, the expression for the new
equilibrium constant will be K = ....... (2)
The magnitude of the equilibrium constant is not affected by the changes in concentrations of
reactants and products
Thus, from (1) and (2), we get K = XY2 =

35) HA + OH– A– + H2O

Keq = = =

36)

2NH4Cl + Ba(OH)2 → BaCl2 + 2NH4OH

0.6 m. 500 ml 0.1 m. 500 ml
300 mmol 50 mmol
(4R)
200 mmol 0

pOH = pKb + log = 5 + log

pH = 14 - 5.3 = 8.7
37) gm equivalent of oxidising agent = gm equivalent of KI
moles × n factor = gm eq. of KI = 0.1 × 1

moles ∝

38)

When a particular oxidation state becomes less stable relative to other oxidation state, one
lower, one higher, it is said to undergo disproportionation.
Cu+ → Cu2+ + Cu

39)

2ey + 2H⊕(aq) —→ H2(g)

E=0–
Potential will be negative if
That is satisfied by option (A)

40) N = =

M=
Ksp = s2 = 10–8M2

41) A → B
a 0
a–x x
X = Kt = 0.025 × 40 = 1M
∴ a – x = 2 – 1 = 1M

42) Due to low solubility of Na2SO410H2O its seperate form Na2Cr2O7 by crystallisation,
process.

43)

Negative charge on metal cation ∝ Synergic Bonding

Synergic Bonding ∝ M – C Bond Strength

∴
44) MgSO4. 7H2O Mg3(PO4)2
Δ Co (NO3)2
MgO.CoO + SO2 + NO2 + O2 + H2O
pink

45)

46) ΔHº = 2 × ΔHºf (C) + ΔHºf(B) – 2 × ΔHºf(A) = –290 kJ/mol

Δsº = +2 – = 250 J/K
ΔGº = ΔHº – TΔSº = –540 kJ
–RT ln kp = ΔGº
ln kp = 67.5

47)
n.f = 3 n.f = 5
molar ratio 5 mole : 3 mole

48)

Similarly
 x = a(1 – e–kt)

49)

50) (i) Ag2SO3 + H+ 2 Ag+ + SO2 + H2O

(ii) FeSO4 solution absorbs NO and becomes brown because of the formation fo [Fe(NO)]2+
SO42–.
(iii) Being covalent do not dissociate.
(iv) Ammonium hydroxide in presence of ammonium chloride is group reagent of IIIrd group.
(v) 2[Cr(OH)4]– (green colouration) + 3H2O2 + 2OH– 2CrO42– + 8H2O.

PART 3 : MATHEMATICS

51)

52)

x ± 1 but x = –1 is rejected.
53)

(x2 – 1)2 – (2x2 – 5)2 = 0

(3x2 – 6) (x2 – 4) = 0
and x = ±2
but x = ±2 is rejected

54) 6[x] – [x]2 – 5 > 0

([x] – 1) ([x] – 5) < 0
⇒ x ∈ [2,5)
and sin(x – 1) > 0
so integral values of x are 2,3 and 4.

55)
1 1 2 2 2 3 3 3 3
( c0 + c1) + ( c0 + c1 + c2) + ( c0 + c1 + c2 + c3) + .............
k = 2 + 22 + 23 + ...... + 22015

k= = 22016 – 2

56) (1 + x)100 (1 + x + x2 + .... + x100)

coefficient of

57) coefficients of ym . in exy (ey–1)m

= coefficient of ym in

= coefficient of ym in exy

= coefficient of ym in
= coefficient of ym in

=
=

58) Diff. w.r.t. ‘x’

⇒ 15(3x3 – 4x2 + 5x – 3)14 × (9x2 – 8x + 5)
and put x = 1
⇒ 15 × 1 × 6 = 90

59) xi = number of times i is used

where i = 1, 2, ....., 9
⇒ x1 + x2 + x3 + ..... + x9 = 5
⇒ Number of solution of this equation
= 13C5 = 1287
Number of required numbers
= 2 × 1287 – 9 = 2565

60) Ans. (D)

Sol. x1 W x2 W x3 W x4
x2, x3 ≥ 2, x1, x4 ≥ 0
x1 + x2 + x3 + x4 = 15
Total integral solutions of (x1, x2, x3, x4)
= 14C3
Total arrangements = 14C3

61)

No of (rectangles + squares)

and no. of squares

No of pure rectangles = 1296 – 204

= 1092

∴ ratio

62) Using inclusion and exclusion principle

Required Number of ways = Total number of ways

– + when R3 is empty – when R1 & R2 both empty 3

63)

∴ R is not reflexive

∴ R is symmetric

∴ R is not transitive.

64) Let

(Which is not irrational)

⇒ Not reflexive

but
⇒ Not symmetric.

but
⇒ Not transitive

65) The given relation may be written in set builder form as

R = {(a, b) : a – b divided by n and a, b ∈ Z}
As a – a = 0 and 0 divided by n
∴ (a, a) ∈ R
∴ R is reflexive
Let a, b ∈ Z such that (a, b) ∈ R
Then (a, b) ∈ R ⇒ a – b divided by n.
a – b = nk for some integer k
⇒ b – a = n(– k)
∴ (a, b) ∈ R ⇒ (b, a) ∈ R
∴ R is symmetric
Now, (a, b), (b, c) ∈ R
Now, a – b = nc1 and b – c = nc2 for some integers c1 and c2.
∴ (a – b) + (b – c) = n(c1 + c2)
⇒ a – c = nk, where k = c1 + c2, an integer.
⇒ (a, c) ∈ R.
∴ (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive and hence R is an equivalence relation.

66)
Now by definition, R is Reflexive, Symmetric and Transitive.
67)

B – C = 90°
∴ B + C = 150°
B = 120° C = 30°

68) 27Δ2 = s4
27 s(s – a) (s – b) (s – c) = s4
3[(s – a) (s – b) (s – c)]1/3 = s

≥ [(s – a) (s – b) (s – c)]1/3
∴s–a=s–b=s–c
∴a=b=c

∴ a = 2R sinA = 2 × 4 ×

Area =

69)

70)
∴

Where

Now,

71)

⇒
Hence, x = 0, –1 are the only real solutions.

72) The number of ways of making 30 in 7 shots is the coefficient of

So, coefficient of
73) ;0<f<1
observe, 0 < 2 –
So, let where 0 < f ' < 1
Now, Add,

= 2(32 + 10 × 8 × 3 + 5 × 2 × 9)
[P] + f + f ' = 2 × Integer = 2 × [32 + 240 + 90] = 724
Observe, 0 < f + f ' < 2
& f + f ' = 2 × Integer – [P] = integer
So, f + f ' = 1 ⇒ 1 – f = f '
∴ Note: f ' · P = (2 – )5
⇒ (1 – f) ([P] + f) = 1 ⇒ (1 – f) [P] + f – f2 = 1

⇒ (1 – f)[P] = 1 – f + f2 ⇒ [P] = 1 +

∴ = [P] – 1 = (724 – 1) – 1 = 722

74)

x = 4k + 3
∴ (2020 + x)2022 = (2020 + 4k + 3)2022
= (4(505 + k) + 3)2022
= (4λ + 3)2022 = (16λ2 + 24λ + 9)1011
= (8(2λ2 + 3λ + 1) + 1)1011
= (8p + 1)1011
∴ Remainder when divided by 8 = 1

75)

Given : A, B, C are in A.P.

⇒ 2B = A + C .. (1)
∵ A + B + C = 180° .. (2)
from (1) & (2) ⇒ 3B = 180° ⇒ B = 60°

from sine rule

= 2(sinAcosC + cosAsinC)

E = 2sin(A + C) = 2sinB =
⇒