06-07-2025

1001CJA103096250001 JA

PART-1 : PHYSICS

SECTION-I

1) Two cars moving towards each other having 30 km/hr and 40 km/hr constant speed. Initially
distance between cars is 70 km then time after which they meet, is

(A) 1 sec
(B) 1 hour
(C) 60 sec
(D) 120 sec

2) A ball is projected from a point on the long inclined plane (having angle 60° from horizontal) with
a speed 5m/s which is perpendicular to the inclined plane. The range of the ball on inclined plane
will be :

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

3) Assertion : Newton’s 2nd law of motion gives measurement of the force.

Reason : According to Newton’s 2nd law of motion force is directly proportion to the rate of change
of momentum.

(A) Assertion is True, Reason is True ; Reason is a correct explanation for Assertion.
(B) Assertion is True, Reason is True ; Reason is not a correct explanation for Assertion.
(C) Assertion is True, Reason is False.
(D) Assertion is False, Reason is True.

4) A ball is projected up on an inclined plane. If t1 and t2 are time taken from A to B and B to C then
which of the following is wrong : (B is at maximum distance from incline plane)
(A) t1 = t2
(B) x1 = x2
(C) v1 > v3
(D)

5) For the system shown in figure if velocity of B at certain instant is 2m/s in downward direction,

then velocity of A at this instant is-

(A) 3 m/s in downward direction

(B) 3 m/s in upward direction
(C) 6 m/s in upward direction

(D)
m/s in upward direction

6) A cylinder of mass 10 kg is at rest between two frictionless inclined surfaces AB and AC. It is
connected with a vertical rope PQ where other end is tied to roof. If cylinder exerts 30N and 40N
forces on surfaces AC and AB respectively, then tension in string is [g = 10 m/s2]

(A) 100 N
(B) 50 N
(C) 30 N
(D) zero

7) Block A is moving horizontally with speed 5 m/sec, then the speed of block B at the given instant
is

(A) 3 m/sec
(B) 4 m/sec
(C) 2.4 m/sec
(D) 5 m/sec

8) A stone is thrown upward with initial speed u if retardation due to air resistance is bυ2 (where υ is
instantaneous speed) b is any constant. Find speed of stone when it come back on ground (g is acc
due to gravity)

(A)

(B)

(C)

(D) u

9)

In shown situation elevator is moving upward with acceleration of 5 m/s2.

List–I List–II

(I) Net force acting on B (P) 350 N

(II) Normal reaction between A and B (Q) 300 N

 (III) Normal reaction between B and C (R) 200 N

(IV) Normal reaction between C and elevator (S) 750 N

(T) 1500 N

(U) 150 N

If mA = 10 kg, mB = 40 kg, mC = 50 kg then choose the correct match.

(A) (I) - (R) ; (II) - (U) ; (III) - (S) ; (IV) - (T)
(B) (I) - (R) ; (II) - (U) ; (III) - (T) ; (IV) - (S)
(C) (I) - (R) ; (II) - (S) ; (III) - (U) ; (IV) - (T)
(D) (I) - (R) ; (II) - (T) ; (III) - (S) ; (IV) - (U)

10) If point A of rod is moving with velocity v0 in right direction then speed of B is :

(A) v0tanθ
(B) v0cotθ

(C)

(D)

11) Choose the incorrect option :-

If relative velocity of one object with respect to other is towards the line joining them, then they
(A)
must collide.
If relative velocity of one object with respect to other is towards the line joining them, then they
(B)
may collide.
A particle starting from rest under the action of constant acceleration must follow a straight
(C)
line trajectory.
Motion under constant acceleration can never result in a closed path trajectory with some area
(D)
enclosed

12) Two cars start from same position with acceleration 5 m/s2 & 10 m/s2 in opposite direction. What
will be the separation between them at t = 4 sec?

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

13) A particle is projected from point O on the ground with velocity at angle
 . It strikes at point C on a fixed smooth plank AB having inclination 37°with
horizontal as shown in figure. Find the x - coordinate of point C in metres. ( Given that

(A)

(B)

(C) 5
(D) 10

14) During a projectile motion, if the maximum height equals the horizontal range, then the angle of
projection with the horizontal is

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

15) Statement–I : Path of a projectile projected at an acute angle to horizontal is a parabola

irrespective of its velocity of projection. because
Statement–II : Trajectory of a projectile projected at an acute angle to horizontal is a parabola
when variation of g, (acceleration due to gravity) can be neglected.

(A) Statement–I is true, Statement–II is true ; Statement–II is correct explanation for Statement–I.
Statement–I is true, Statement–II is true ; Statement–II is NOT a correct explanation for
(B)
statement–I.
(C) Statement–I is true, Statement–II is false.
(D) Statement–I is false, Statement–II is true.

16) Suppose a player hits several baseballs. Which baseball will be in the air for the longest time?

(A) The one with the farthest range.

(B) The one which reaches maximum height.
(C) The one with the greatest initial velocity.
(D) The one leaving the bat at 45° with respect to the ground.

17) The motion of a particle is defined by the position vector = A(cos t + t sin t) + A (sin t – t cos
t)
Where t is expressed in seconds. Determine the value of t for which positions vectors and
acceleration vector are perpendicular.
(A) 0
(B) 1
(C) 3
(D) 5

18) A ball is hit by a batsman at an angle of 37° as shown in figure. The man standing at P should
run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that
height of man is negligible in comparison to maximum height of projectile.

(A) 3 m/s
(B) 5 m/s
(C) 9 m/s
(D) 12 m/s

19) Mandy stands on a weighing scale inside a lift (elevator) that accelerates vertically upwards as
shown in the diagram below. The forces on Mandy are her weight W and the reaction force from the

scale R. The reading of the scale is :

(A) R + W
(B) W
(C) R
(D) R – W

20) An object is projected up the incline at the angle 37° with incline as shown in figure with an
initial velocity of 20 ms–1. The angle of inclination of inclined with horizontal is also 37°. Find the
range (in m) of particle along inclined surface. (g = 10 m/s2)

(A) 21
(B) 7
(C) 14
(D) 28

SECTION-II

1) How much force is required to move 20 kg block with 5 m/s2 acceleration (in N) ?

2) In the figure shown C is a fixed wedge. A block B is kept on the inclined surface of the wedge C.
Another block A is inserted in a slot in the block B as shown in figure. A light inextensible string
passes over a light pulley which is fixed to the block B through a light rod. One end of the string is
fixed and other end of the string is fixed to A.S is a fixed support on the wedge. All the surfaces are

smooth. Masses of A and B are same. Then the magnitude of acceleration of A is . Then x is

(sin 37º = 3/5)

3) The point of suspension of the pendulum bob moves with a constant horizontal acceleration a. If
breaking strength of the string is equal to three times the weight of the bob, then maximum value of

a is . Find x.

4) The velocity of a body of mass 2 kg changes

from to in 3 sec, then the magnitude of the average force applied: (in N) (Round off to the Nearest
Integer)

5) Two monkeys each of mass m move with acceleration relative to the light inextensible
sting as shown in the figure. The ratio of tensions in the portions AB and BC of the string is :-
 PART-2 : CHEMISTRY

SECTION-I

1) Consider the following orders :

(I) Atomic size : In > TI > Al > Ga > B
(II) Ionization enthalpy : B > TI > Ga > Al > In
(III) Electron affinity : N > P > As > Sb
(IV) E.N. of central atom : C2H2 > C2H4 > C2H6
The correct orders are :

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

2) Statement -1 : Two successive ionisation energies of Argon are 56.8 eV and 36.8 eV respectively.
because
Statement -2 : Zeff of Ar (3s23p6) is greater than Ar+ (3s23p5).

(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for
(B)
statement-I
(C) Statement-I is true, Statement-II is false
(D) Statement-I is false, Statement-II is false

3) From each pair given below, identify the ion which is larger in size.
(Li+, Mg2+), (Cu2+, Zn2+),(Na+, F–), (Ce3+, Lu3+)

(A) Mg2+, Zn2+, F–, Ce3+

(B) Li+, Zn2+, F–, Ce3+
(C) Mg2+, Cu2+, F–, Lu3+
(D) Li+, Cu2+, F–, Lu3+

4) Choose the incorrect statement(s) about Na, Cs, Cl, N

(A) 'Cs' is more metallic than Na but Na is less electropositive than Cs
(B) 'N' has three electrons in p-subshell but 'Cℓ' has elevan
Anion of 'Cl' is having largest size than cation of 'Na' but cation of 'Na' has smaller size than
(C)
cation of 'Cs'
(D) Except Cl all have +ve ΔHEGE

5) Which of the following Ist ionisation energy order is/are correct

(A) Be < B < C

(B) N < O < F
(C) Li > Be > Cs
(D) P > S > Ca

6) Which of the following order is correct according to ionisation energy ?

(A) Sc < Y < La

(B) Ti > Zr > Hf
(C) Mo > Cr > W
(D) Pt > Pd > Ni

7) Which of the following order for given property is WRONG -

(A) HF < HCl < HBr (Internuclear distance)

(B) Mg < S < Mn (valence electron)
(C) Ni < Pt < W (total number of unpaired electron)
(D) Zn+2 < Cd+2 < Hg+2 (Ionic radii)

8) Aspartame, an artificial sweetener contains 9.52 wt.% nitrogen. There are two nitrogen atoms per
molecule. What is the molecular weight of aspartame ?

(A) 147
(B) 294
(C) 588
(D) 266

9) The empirical formula of a compound is CH2O. If 0.0833 moles of the compound contains 1.0 g of
hydrogen, then its molecular formula should be

(A) C6H12O6
(B) C5H10O5
(C) C4H8O4
(D) C3H6O3

10) Assertion : Specific gravity is dimensionless.

Reason : Specific gravity is density of substance measured w.r.t. density of water.
(A) If both the Assertion and Reason are True and Reason is the correct explanation of Assertion.
(B) If both the statement are True but Reason is not the correct explanation of Assertion.
(C) If Assertion is True and Reason is False.
(D) If Assertion is False and Reason is True.

11) If H2SO4 is formed from it’s elements by taking 6.022 × 1023 atom of ‘O’ 5.6 litre of H2 gas at 0°C,
1 atm and 8 g S then the correct information is :

(A) 0.125 moles of H2SO4 are formed

(B) 0.25 moles of H2SO4 are formed

(C)
mole of H2 is left
(D) 1/4 mole of O2 is left

12) The total mass of a mixture containing 5.6 L of H2O (l, density = 1gm/ml) and 1 g atoms of Zn is
:- [H = 1, O = 16, Zn = 65.5]

(A) 70 g
(B) 71.1 g
(C) 74.5 g
(D) 5665.5 g

13)

From 2 mg calcium, 1.2 × 1019 atoms are removed. The number of g-atoms of calcium left is (Ca =
40, NA = 6 × 1023 )

(A) 5 × 10–5
(B) 2 × 10–5
(C) 3 × 10–5
(D) 5 × 10–6

14) Which of the following is correct regarding mention property in bracket?

(A)
(Stability of carbocation)
(B) CH3–CH=CH2 < MeO–CH=CH2 (Rotational barrier energy)

(C)
(Carbon oxygen bond length)
(D)

(Resonance energy)

15) Which of the following species is/are aromatic?

(A) (I) and (II)

(B) (II), (III) and (IV)
(C) (I) and (IV)
(D) (I) and (III)

16)
Which of the following arrangement gives the correct order of decreasing stability of the above
mentioned resonance contributions.

(A) I > II > IV > III

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

17) The correct order of electron density in aromatic ring of following compound is-

(A) I > III > II > IV

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

18) Which of the following isomeric hydrocarbon is most acidic?

(A)

(B)

(C)

(D)

19) Decreasing order of acidic strength is

(A) II > I > III > IV

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

20) Matching List :-

List-I List-II

(P) (1) +M

(Q) (2) –M

(R) –OH (3) +I

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

SECTION-II

1) The number of elements possible in second period of periodic table if azimuthal quantum number
(l) can have integral values from 0,1,2.......................(n+1) is_____.

2)

N2O4 dissociates into NO2, if % dissociation of N2O4 is 33.33 %. Calculate average molecular weight
of gaseous mixture formed. [Atomic weight : N = 14, O = 16]

3) 5.00 moles of hydrogen gas, 3 moles of white phosphorus {P4(s)} and 12 moles of oxygen gas are
taken in a sealed flask and allowed to react as follows : H2(g) + P4(s) + O2(g) —→ H3PO4 Determine
the moles of ortho-phosphoric acid that can be produced, considering that the reaction occurs in
90% yield.

4) The total number of contributing structures showing hyperconjugation (involving C–H bonds) for

the following carbocation is

5) The number of resonance structures for M is :

M⇒

PART-3 : MATHEMATICS

SECTION-I

1) If α, β are the integral roots of the quadratic equation x2 – αβx + α2 + β = 0, then number of
possible different quadratic equation(s) is -
(A) 0
(B) 1
(C) 2
(D) 4

2) If common ratio of a geometric progression is positive then maximum value of the ratio of second
term to the sum of first 3 terms equals -

(A) 2

(B)

(C) 3

(D)

3) Assertion (A) : If the equation ax2 + bx + c = 0 (a,b,c ∈ R – {0}) and 4x2 + 5x + 6 = 0 have a

common root then is .

Reason (R) : Roots of 4x2 + 5x + 6 = 0 are imaginary so both the roots are common.

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

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

4) The minimum value of , where x and y are positive real numbers, is -

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

5) If a, b, c are integers and b2 = 4(ac + 5d2), d ∈ N, then roots of the equation ax2 + bx + c = 0 are

(A) Irrational
(B) Rational & different
(C) Complex conjugate
(D) Rational & equal

6) If α2 = 5α – 3, β2 = 5β – 3 then the value of is -

(A)

(B)
(C)

(D)

7) If the roots of the equation x2 – bx + c = 0 be two consecutive integers, then b2 – 4c equals -

(A) 1
(B) 2
(C) 3
(D) –2

8) If x2 – A(x + 1) + C = 0 has roots x1 & x2, then the value of x12 + x22 + (2 + A)x1x2, is -

(A) AC
(B) A2 + AC
(C) A2 – AC
(D) –AC

9)

List-I List-II

The value of xyz is 15/2 or 18/5 according as the

(A) series a, x, y, z, b are in A.P. or H.P. then 'a + b' (I) 2
equals where a, b are positive integers.

(B) The value of ....∞ is equal to (II) 1

If x, y, z are in A.P. and

(C) (x + 2y – z) (2y + z – x) (z + x – y) = kxyz, (III) 3
where k ∈ N, then k is equal to

There are m A.M. between 1 and 31. If the ratio

(D) (IV) 4
of the 7th and (m – 1)th means is 5 : 9, then is
equal to
(A) A → IV, B → I, C → IV, D → I
(B) A → IV, B → I, C → IV, D → II
(C) A → II, B → I, C → IV, D → II
(D) A → I, B → II, C → III, D → IV

10) If α, β and γ are three consecutive terms of a non-constant G.P. such that the quadratic
equations αx2 + 2βx + γ = 0 and x2 + x – 1 = 0 have a common root, then α(β + γ) is equal to (where
α, β, γ ∈ R)

(A) βγ
(B) 0
(C) αγ
(D) αβ

11) Let α and β be the roots of the equation 5x2 + 6x – 2 = 0. If Sn = αn + βn, n = 1, 2, 3 ..., then :

(A) 5S6 + 6S5 = 2S4

(B) 5S6 + 6S5 + 2S4 = 0
(C) 6S6 + 5S5 + 2S4 = 0
(D) 6S6 + 5S5 = 2S4

12) If x ∈ R, then number of integers in the range of are

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

13) If the roots of equation 4x2 + 4ax + b = 0 are real and differ at most by 'a' (a > 0), then complete
range of values in which b lies in-

(A)

(B)

(C)

(D)

14) If a1, a2, a3, ...., a40 are in A.P. and and then a40 equals :

(A) 110
(B) 115
(C) 120
(D) 125

15) Statement-I : If x2 + 2ax + 10 – 3a > 0 for each x ∈ R, then value of a is –5 < a < 2.
Statement-II : If px2 + qx + r > 0 ∀ x ∈ R (where p, q, r ∈ R & p ≠ 0), then q2 < 4pr.

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

(B) Both Statement-I and Statement-II are false.
(C) Both Statement-I and Statement-II are true.
(D) Statement-I is false but Statement-II is true
16) If (20)19 + 2(21) (20)18 + 3(21)2 (20)17 + ... + 20(21)19 = k(20)19 then k is equal to

(A) 400
(B) 100
(C) 441
(D) 420

17) If the product of three terms of G.P. is 512. If 8 added to first and 6 added to second term, so
that number are in A.P., then the numbers are

(A) 2, 4, 8
(B) 4, 8, 16
(C) 3, 6, 12
(D) 1, 2, 3

18) Sum of series ..........

(A)

(B)

(C)

(D)

19) Let 3 geometric means G1, G2, G3 are inserted between two positive number a and b such that

= 2, then equal to

(A) 2
(B) 4
(C) 8
(D) 16

20) Let the number of terms common to the two APs, 11, 15, 19, ..., (upto 53 terms) and 4, 9, 14, ...

(upto 44 terms) be λ, then is

(A) 3
(B) 5
(C) 7
(D) 9

SECTION-II
1) If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 + px + q = 0 has equal
roots, then the value of '4q' is -

2) If the 2nd, 5th and 9th terms of a non-zero & non-constant A.P. are in G.P. & let the common ratio of
this G.P. is 'r' then '18r' is :-

3) If abcd = 1 where a, b, c, d are positive real then the minimum value of a2 + b2 + c2 + d2 + ab + ac

+ ad + bc + bd + cd is

4) For all x, x2 + λx + > 0 then the number of non-positive integral value(s) of λ is(are)

5) Let f(x) = ax2 + bx + c, where a ≠ 0, a, b, c are integers and f(1) = 1, 6 < f(3) < 8
and 18 < f(5) < 22. Then the number of solutions of equation f(x) = ex 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. B B A B B B A C A B A C C D D B B B C A

SECTION-II

Q. 21 22 23 24 25
A. 100 32 8 4 4

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

SECTION-II

Q. 46 47 48 49 50
A. 18 69 3 7 5

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

SECTION-II

Q. 71 72 73 74 75
A. 49 24 10 10 1
 SOLUTIONS

PART-1 : PHYSICS

1) vrel = 30 + 40 = 70 km/h

time = = 1 hour.

2)

ux = 0 , uy = 5m/s, ay = 5, ax =

Time of flight =
∴ Range on inclined plane

3)

Theoretical Question

4) Horizontal component of velocity is decreasing with time.

5) Let tension in string be T then (2T)vA = (3T)vB ⇒ vA = × 2 = 3 ms–1 in upward direction.

6)
Resultant of 30 N & 40 N is 50 N in vertical direction. Hence T = 50 N
7)
String constraint
= constant

= –cosθ × VA

= – cos 53° × 5 = – = – 3 m/sec

8)

upward motion = –(g + bv2)

ln = 2bH

downward motion : = g – bv2

ln = 2bH

=
9)

For (I) : Net force on B = ma = (40) (5) = 200N

For (II) : FBD of A

NAB – 10 g = 10(5) ⇒NAB = 150N

For (III) : FBD of B

NBC –NAB – 40 g = 40(5) ⇒NBC = 750N
For (IV) : For system of A + B + C :
NC – 100 g = 100 (5)⇒ NC = 1500 N

10)
v0cosθ = v1sinθ
v1 = v0cotθ

11) Relative velocity must be towards the observer along the line joining.

12)

arelative = 15 m/s2

= 120 m

13)
Solving the equation gets

14) R = Hmax

tanθ = 4
θ = tan(4)

15) Trajectory of a projectile is a parabola when variation of g, (acceleration due to gravity)

can be neglected, so if initial velocity is very high then g may vary and its trajectory will not be
parabola.

16)

H ∝ T ∝ uy
∴ Ball with maximum height have maximum time of flight.

17)
and
⇒ t = 1 sec

18) ux = 15 × = 20 m/s

uy = 15 × = 15 m/s

umb = um – ub
20 = um – 15
um = 5 m/s

19)
Reaction force from scale is the reading of scale.

20)

= 48 – 27 = 21 m

21)
Tension is internal force as no effect on net acceleration.

22) Along the incline :-

2mgsin37° – T = 2ma
Perpendicular to incline,
T – mgcos37° = mb
Constraint equation :-
a=b

∴a=b=

∴ anet of A =

23)
T sin θ = ma ; T cos θ = mg
∴ T2 = (mg)2 + (ma)2

9g2 = g2 + a2 ⇒ a2 = 8g2
a= g
∴ x=8
 24)

25)

PART-2 : CHEMISTRY

26) At. Size : B < Ga < Aℓ < In < Tℓ

E.A. : N < P < As < Sb

27) Successive IE, increases in magnitude

28)

Li+ > Mg+2

Li+⇒ 76 Pm Cu+2 ⇒ 73 Pm
+2 +2 Cu+2 < Zn+2
Mg ⇒ 72 Pm Zn ⇒ 75 Pm
Na+ < F–, Ce+3 > Lu+3

29) Cs is most metallic

N has three electrons in 2p-subshell
Anion of Cl is having largest size than cation of Na
Cl, Na, Cs have –ve EGE

30) Let's analyze each option for the ionization energy order :
The species with the lowest nuclear charge and hence the lowest ionization energy and I.E.
also depends on size.

31) Theory based

32) Ni ⇒ [Ar] 3d8 4s2

Pt ⇒ [Xe] 4f14 5d9 6s1
W ⇒ [Xe] 4f14 5d4 6s2
33) Molecular weight of aspartame

= × 14 × 2 = 294 Ans.]

34) Let mol. formula be CxH2xOx

0.0833 × 2x = 1 ⇒ x = 6

35)

Specific gravity is unitless. Because it is defined density of water ( = 1 gm/ml)

36) H2 + S + 2O2 → H2SO4

= = ns = = =
As all reactants are in stoichiometric ratios, none will be left behind.

Hence mole of H2SO4 is formed.

37) H2O is not a gas at NTP.

Use density of H2O = 1 g ml–1

38)

39) (1) Stability of carbocation ∝ Resonance

(2) Rotational barrier energy ∝ Double bond character
(4) Phenoxide ion does not have change separation

40) (I) have in complete cyclic conjugation so aromatic in

nature.

(II) annulene [10] due to repulsion between H-atom its planarity

disturbed and non aromatic in nature.
(III) Complete cyclic conjugation not present so non aromatic.

(IV) Complete cyclic conjugation and so aromatic in nature.

41) (I) and (IV) have maximum number of covalent bonds but non polar is more stable than
polar. (III) is least stable resonating structure because negative and lone pair on adjacent
atom.

42) On the base of +m and –m order and hyperconjugation

> > >

43)

44) Acidic strength ∝ Stability of conjugate base.

 (A) is more stable than (B) therefore conjugate acid of (A) is more acidic.

45) (P) is +I
(Q) is –M, –I
(R) is +M, –I
(S) is –I

46) 2s, 1d, 2p are filled

47) Meff =

1 + 1/3 =
Mav = 69

48)

6H2(g) + P4(g) + 8 O2(g) 4H3PO4

Mole 5 mol 3 mole 12 mole
L.R.

Mole of H3PO4 =

49)

The correct answer is 7

50)

The correct answer is (5)

PART-3 : MATHEMATICS

51) α + β = αβ
⇒ α + β – αβ – 1
⇒ (α – 1) (1 – β) = –1
∴ α = 0 = β or α = 2 = β
but αβ = α2 + β ⇒ α = 0 = β
⇒ only one quadratic equation

52) Let the first 3 terms be , a, ar

Hence

53) 4x2 + 5x + 6 = 0 ...(i)

ax2 + bx + c = 0 ...(ii)

since D = 25 – 4.4.6 < 0

⇒ a = 4k, so equation (1) has
b = 5k imaginary roots
c = 6k

so

54) AM ≥ GM ⇒

55)

b2 – 4ac = 20 d2

56) α, β are the roots of the equation x2 – 5x + 3 = 0

α + β = 5 and αβ = 3
⇒ α2 + β2 + 2αβ = 25
⇒ α2 + β2 = 19

57) If roots are two consecutive integers

⇒ say I and I + 1 then
difference of roots = 1

58) x2 – A(x + 1) + C = 0

x2 – Ax + C – A = 0
⇒ x1 + x2 = A ....(i)
x1x2 = C – A ....(ii)
2 2 2
x1 + x2 + (2 + A)x1x2 = (x1 + x2) + Ax1x2
= A2 + A(C – A)
= AC

59)

(A)

(a + d) (a + 2d) (a + 3d) =

(b + 3a) (a + b) (a + 3b) =
= 15×16 .............(1)

in A.P.

(ab)3 = 9×3
ab = 3
then, a = 1, b = 3
a+b=4
(B)
Let

–
_____________________

s =1
= 21 = 2
(C) (x + 2y – z)(2y + z – x)(z + x –y) = kxyz
(x + x + z – z)⋅(x + z + z – x)(2y – 4) = kxyz
2x⋅2z⋅y = kxyz
k=4

(D)
9 + 63d = 155 – 10d ⇒ 73d = 146

m + 1 = 15 ⇒ m = 14 ⇒

60) αx2 + 2βx + γ = 0

Let β = αt, γ = αt2
∴ αx2 + 2αtx + αt2 = 0
⇒ x2 + 2tx + t2 = 0
⇒ (x + t)2 = 0
⇒ x = –t
it must be root of equation x2 + x – 1 = 0
∴ t2 – t – 1 = 0 ....(1)
Now
α(β + γ) = α2(t + t2)
Option 1 βγ = αt . αt2 = α2 t3 = a2 (t2 + t)
(from equation 1)

61) α and β are roots of 5x2 + 6x – 2 = 0

⇒ 5α2 + 6α – 2 = 0
⇒ 5αn + 2 + 6αn + 1 – 2αn = 0 ....(1)
(By multiplying αn)
Similarly 5βn + 2 + 6βn + 1 – 2βn = 0 ....(2)
By adding (1) & (2)
5Sn + 2 + 6Sn + 1 – 2Sn = 0
For n = 4

62)
⇒ x2(y – 1) + 2x(y – 1) + 7y – 1 = 0
Clearly y ≠ 1
Now, D ≥ 0
⇒ 4(y – 1)2 – 4(y – 1)(7y – 1) ≥ 0
(y – 1)(y – 1 – 7y + 1) ≥ 0
y(y – 1) ≤0 ⇒ y ∈ [0, 1]
⇒y=0

63) for real roots D > 0

16a2 – 16b > 0
∴ b < a2
Also |α – β| < a
(α + β)2 – 4αβ < a2
b>0
∴ 0 < b < a2

64)

a2 + a4 + a6 + .... + a40 = 400

a + d + 19d = 20
a + 20d = 20 ..... (1)
a1 + a3 + a5 + ..... + a39 = 300

a + 19d = 15 .... (2)

Solving (1) & (2) we get
d = 5, a = –80
a40 = –80 + 39 × 5 = 115

65) x2 + 2ax + 10 – 3a > 0 ∀ x

⇒ 4a2 – 4(10 – 3a) < 0
a2 + 3a – 10 < 0
a2 + 5a – 2a – 10 < 0
a(a + 5) –2(a + 5) < 0
(a + 5) (a – 2) < 0

–5 < a < 2

66) Dividing by (20)19, we can write k = 1 + 2x + 3x2 + ... + 20x19

where x = 21/20.

67) Let three terms of G.P. are , a, ar

.a.ar = 512
3
a = 512
a=8

Given : , (a + 6), ar ......... A.P.

So 2(a + 6) = + 8 + ar
Put the value of a

2(8 + 6) = + 8 + 8r

20 = + 8r

5 = + 2r
2r2 – 5r + 2 = 0
r = 2,

For r = 2 For r =

, 8, 8 × 2 , 8, 8 ×
= 4, 8, 16 16, 8, 4
Ans. (B)

68)

Ans. (A)

69) If a, G1, G2, G3, b are in G.P. with common ratio equal to 'r' then G1– a, G2 –G1, G3 – G2, b–G3

are also in G.P. with same common ratio ⇒ =2⇒ = r4 = 16

70) Obviously 11 terms are common.

71) One root of x2 + px + 12 = 0 is 4

⇒ (4)2 + p(4) + 12 = 0
⇒ p = –7
x2 + px + q = 0 has equal roots
⇒ D = p2 – 4q = 0
⇒ 49 – 4q = 0

72) t2 = a + d
t5 = a + 4d
t9 = a + 8d
Given t2, t5, t9 are in G.P.
(a + 4d)2 = (a + d) (a + 8d)
a2 + 16d2 + 8ad = a2 + 8d2 + 9ad
8d2 – ad = 0
d(8d – a) = 0
As given non-constant AP. ⇒ d ≠ 0

∴ d = ⇒ a = 8d
so, A.P. is 8d, 9d, 10d, ...

Common ratio of G.P. =

73) AM ≥ GM

a2 + b2 + c2 + d2 + ab + ac + ad + bc + bd + cd ≥ 10
minimum value = 10

74) Use : D < 0 ⇒

⇒ λ2 + 6λ – 40 < 0
⇒ (λ + 10)(λ – 4) < 0
⇒ λ ∈ (–10, 4)

75)

Given a + b + c = 1
9a + 3b + c = 7
18 < 25a + 5b + c < 22
⇒ From above (1), (2) and (3)
4 < 7a– b – c < 8
4 < 7a – b + a + b – 1 < 8
5 < 8a < 9

a = 1, b = –1, c = 1
ƒ(x) = x2 – x + 1