22-06-2025

0999DJA161101250001 JA

PART-1 : PHYSICS

SECTION-I (i)

1) Find value of [1 – cos 1.8°] using approximation.

(A)

(B)

(C)

(D)

2) Precession is the result of the angular velocity of rotation (spin) and the angular velocity produced
by the torque. It is an angular velocity about a line that makes an angle θ with the permanent
rotation axis. In the diagram, a top of mass m is performing precession motion. The torque due to
gravitational force Fg causes a change in the angular momentum L in the direction of that torque
causing the top to process with angular velocity ωP which is given by: (Given r is the distance of
COM from point of rotation ‘P’. Is & ωs are the moment of inertia and angular velocity of top about

spin axis respectively.)

(A)

(B)
(C)

(D)

3) A parallelogram is formed with as sides. Let be the diagonals of the

parallelogram. Then a2 + b2 = .......

(A)
(B)

(C)

(D)

4) A particle is moving westward with a velocity = 5 m/s. Its velocity changed to = 5 m/s

northward. The change in velocity vector is :

(A) m/s towards north east

(B) 5 m/s towards north west
(C) zero
(D) m/s towards north west

SECTION-I (ii)

1) The velocity, acceleration and force in two systems of units are related as under : i. ii.

iii.
All the primed symbols belong to one system and unprinted ones belong to the other system, α and β
are dimensionless constants. Which of the following is/are correct?

(A)
Length standards of the two systems are related by : .

(B)
Mass standards of the two systems are related by :

(C)
Time standards of the two systems are related by :

(D)
Momentum standards of the two systems are related by :

2)
In a new system of units, the unit of mass is 1000 kg (1 metric ton), unit of length is 1000m (1km)
and unit of time is 3600 s(1hr). Select correct statement(s).

(A) The numerical value of 500 m in this new system is 0.5.

(B) The numerical value of 7200 s in this new system is 2.
(C) The numerical value of 1/36 J in this new system is 3.6 × 10–4.
(D) The numerical value of 1/36N in this new system is 0.36.

3) If angle between and is θ (0° < θ < 180°); angle between and is α; angle between
and is β then which of the following option(s) is/are CORRECT ?

(A)
If then 2α > θ

(B)
If then 2α < θ

(C)
If then 2β + θ > 180°

(D)
If then α + β = 90°

4) Select CORRECT statement(s) for three vectors , and

(A) The above vectors can form triangle.

(B) Component of a along is 3

(C)
makes angle with y-axis.
A vector having magnitude twice the vector and anti parallel to vector is
(D)

SECTION-I (iii)

Common Content for Question No. 1 to 2

Abhishek, Divyam, Nitu [Allen students] came to center to collect modules. After collecting them,
they left for their home. Abhishek first goes 1 km towards East direction then takes turn and goes 2
km towards North then 3 km towards West to get to his home. Divyam first goes 2 km North, 4 km
West then 5 km South. Nitu first goes 5 km South then 4 km West and finally 9 km North.

1)

Shortest distance between Abhishek house and Divyam house is

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

2)

If we join the points of location of the houses of all three students we will get a triangle. Area of the
triangle is :-

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

Common Content for Question No. 3 to 4

In ancient time, different system of units was followed. In which unit of mass was "ser", unit of
length was "gaj" & unit of time was "kaal"
1 ser = 900 gm
1 gaj = 90 cm
1 kaal = 3 hr

3) The unit of volume of water in the ancient system will be :-

(A) ser3
(B) gaj3
(C) gaj–3
(D) kaal3

4) A ball of mass 9 kg is moving with velocity 10 m/s. Then its momentum (mv) in ancient system will
be :-

(A) 1.2 × 106 ser gaj–1 kaal

(B) 10–6/1.2 ser gaj kaal–1
(C) 1.2 × 106 ser gaj kaal–1
(D) 120 ser gaj kaal–1

SECTION-III

1) If the velocity of light (c), gravitational constant (G) and the Planck’s constant (h) are selected as

the fundamental units, then the dimentional formula for mass in new system of units is .
Find the value of (α + β + γ) × 2.

2) A toy helicopter made of paper with rotor radius r and weight W is dropped from a height h in air
of density ρ. When dropped, the helicopter falls down, accelerating to a certain velocity (in
negligible time) and then falling with the same constant velocity. The formula for the time it takes to
fall down is . Find the value of a + b + c – d
3) Two particles having position vectors (at t = 0) metres and metres

are moving with constant velocities m/s and m/s respectively. If they
collide after 2 seconds, then the value of 'α' is :

4) Two forces P and Q are in ratio P : Q = 1 : 2. If their resultant is at an angle to

vector P and if angle between P and Q is 10x (in degree). Find x.

5) if are unit vectors and . Then 4x is equal to :-

6) A bird is at a point P (4, –1, –5) and sees two points P1 (–1, –1, 0) and P2 (3 , –1, –3). At time t = 0,
it starts flying with a constant speed of 10 m/s to be in line with points P1 and P2 in minimum
possible time t. If value of "t" (in sec.) is 100n. Find "n". All coordinates are in kilometers.

SECTION-IV

1) In a new system (say TK system) of measurement, the fundamental quantities length, mass and
time are measured in Akshay, Shahrukh and Aamir respectively.
1 Akshay = 1 km
1 Shahrukh = 1 Quintal (100 kg)
1 Aamir = 1 minute

Column-I Column-II

(A) One unit of acceleration in TK system. (P)

SI unit

(B) One unit of kinetic energy in TK system (Q)

MKS unit

(C) One unit of pressure in TK system (R)

MKS unit

(D) One unit of work in TK system (S)

MKS unit

(T)
SI unit

2) Answer the following by appropriately matching the columns based on the information given in
the paragraph. In a regular hexagon then express all vector in terms of them :
 Column-I Column-II

(A) (P)

(B) (Q)

(C) (R)

(D) (S)

(T)

PART-2 : CHEMISTRY

SECTION-I (i)

1) The molar mass of normal water is .....as compared to heavy water.

(A) 10% less

(B) 10% high
(C) 2% less
(D) zero % less

2) 3 × 1020 molecules of SO3 are added to 160 mg of SO3, then number of moles of SO3 finally
present:(NA = 6 × 1023) [Atomic weight of S = 32]

(A) 1.5 × 10–3

(B) 3 × 10–3
(C) 2.5 × 10–3
(D) 1.5 × 10–4

3) Calculate the molecular formula of compound which contains 20% Ca and 80% Br by weight, if
molecular weight of compound is 200 ?
(Atomic wt. Ca = 40, Br = 80)

(A) Ca2Br
(B) CaBr2
(C) CaBr
(D) Ca1/2Br

4) 5 mole H2O2(ℓ) is placed in a container of volume 490 ml at 300 K, where it is completely

decomposed into H2O(ℓ) and O2(g). The pressure exerted by O2 gas formed is finally. (Given =
1 g/ml, R = 0.08 atm-L/mol-K)

(A) 150 atm

(B) 122.45 atm
(C) 153.75 atm
(D) 300 atm

SECTION-I (ii)

1) To check the principle of multiple proportions, a series of pure binary compounds (PmQn) were
analyzed and their composition is tabulated below. The correct option(s) is(are)

Compound Weight % of P Weight % of Q

1 50 50
2 44.4 55.6
3 40 60
(A) If empirical formula of compound 3 is P3Q4, then the empirical formula of compound 2 is P3Q5.
If empirical formula of compound 3 is P3Q2 and atomic weight of element P is 20, then the
(B)
atomic weight of Q is 45.
(C) If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is P5Q4.
If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of
(D)
compound 1 is P2Q.

2) If 100 mL of 1 M H2SO4 solution is mixed with 100 mL of 9.8%(w/w) H2SO4 solution (d = 1 g/mL),
then the correct information(s) for the final solution is/are :
(Assume volume to be additive)

(A) Concentration of solution is 1M

(B) Volume of solution become 200 mL
(C) Mass of H2SO4 in the solution is 98 g
(D) Mass of H2SO4 in the solution is 19.6 g

3) A mixture of C3H8(g) & O2 having total volume 100 ml in an eudiometry tube is sparked & it is
observed that a contraction of 45 ml occurred. What can be the composition of reacting mixture ?
Assume only the reaction of complete combustion.

(A) 15 ml C3H8 & 85 ml O2

(B) 25 ml C3H8 & 75 ml O2
(C) 45 ml C3H8 & 55 ml O2
(D) 55 ml C3H8 & 45 ml O2
4) 1 g atom of nitrogen represents :

23
(A) 6.02 × 10 N2 molecules
(B) 22.4 litre of N2 at 1 atm, 273 K
(C) 11.2 litre of N2 at 1 atm, 273 K
(D) 14 g of nitrogen

SECTION-I (iii)

Common Content for Question No. 1 to 2

For a compound, it is given that :

(i) Compound has 2 : 1 'H' to 'O' atoms (numbers of atoms)
(ii) Compound has 40% C by mass

(iii) Approximate molecular mass of the compound is 180 u (iv) Compound contains C, H & O only.

1) What is the % by mass of oxygen in the compound :

(A) 53.33%
(B) 88.88%
(C) 33.33%
(D) none of these

2) What is empirical & molecular formula of the compound :

(A) CH3O & C2H6O2

(B) CH2O2 & C2H4O4
(C) CH2O & C6H12O6
(D) C2H2O & C6H14O12

Common Content for Question No. 3 to 4

2.7 gm of Al is heated with 100 mL of H2SO4 (29.4% w/w, density 1 gm/mL) following reaction takes
place :
Al + H2SO4 → Al2(SO4)3 + H2

3)

The volume of H2 gas evolved at 1 atm & 273K :

(A) 3.36 L
(B) 2.24 L
(C) 4.48 L
(D) 11.2 L

4) The hydrogen gas obtained in the reaction is mixed with 1.2 × 1024 molecules of O2. The average
molecular weight of resulting gaseous mixture is. (Assuming gases are not reacting) (N0 = 6 × 1023)
(A) 29.90
(B) 35.6
(C) 17
(D) 24.8

SECTION-III

1) A container contain 5 mole O3, 6 moles KI and 72g H2O react according to
O3 + 2KI + H2O → I2 + 2KOH + O2
then how many moles of H2O left ?

2) Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a
surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is

3) Assume isotope of chlorine present on the unknown planet are and . If

average molecular weight of Cl is found to be 35. What is the sum of moles of proton and neutron
in 7 gm of sample of chlorine

4) How many blood cells of 5ml each having [K+] = 0.1M should burst into 25 ml of blood
plasma [K+] = 0.02 M so as to give final [K+] = 0.06 M?

5) 35% w/v, 400 ml of NH4OH mixed with 12M, 600 ml of H2SO4.Find the molarity [NH4+] in final
solution.

6) A 20 ml mixture of C2H4 and C2H2 undergoes sparking in gas eudiometer with just sufficient
amount of O2 and shows contraction of 37.5 ml. Volume (in ml) of C2H2 in the mixture is.

SECTION-IV

1)

Column-I Column-II

120 g CH3COOH in 1 L
(A) (P) M=2
solution (dsol = 1.2 g/mL)

120 g glucose dissolved in

(B) (Q) 10% w/w solution
1 L solution (dsol = 1.2 g/mL)

(C) = 1/31 (R) 12% w/v solution

(aqueous solution)

19.6% (w/v) H2SO4 solution

(D) (S) m = 1.85
→ (dsolution = 1.2 g/mL)

(T) m = 0.617
2)

Column-I Column-II

(A) Mass of one molecule of H2SO4 (P) 24

(B) Molar mass of H2SO4 (Q) 98 amu

(C) 1 amu (R) 98 gm

(D) Relative atomic mass of Mg (S) 1.66 × 10–27kg

(T) 1.66 × 10–24 kg

PART-3 : MATHEMATICS

SECTION-I (i)

2 2
1) If α and β are solutions of x – 10x + 10 = 0 such that , then roots of equation 2x + 10x
+ 3b = 0 are

(A) rational and distinct

(B) real and equal
(C) imaginary
(D) irrational and distinct

2) If both roots of x2 + px + q = 0 are positive and one root is cube of other root, then -

(A) q3 – 2q2 – p3 + 4p + q = 0
(B) q3 – 2q2 – p4 + 4p2q + q = 0
(C) q3 – 2q2 – p2 – 4pq + q = 0
(D) q3 – 2q2 – p3 – 4pq + q = 0

3)

Graph of y = ax2 + bx + c is given in the adjacent figure, then which of the following is correct -

(where α,β are the zeros of ax2 + bx + c)

(A) α + β > 0, αβ < 0
(B) b < 0, D < 0
(C) b > 0, D > 0
(D) α + β < 0, αβ < 0

4) If the equations x2 + x + 1 = 0 and x2 – ax + b = 0 have one root common, then 2a + 3b is equal to

(where a,b ∈ R)

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

SECTION-I (ii)

1) The shaded region in the figure is not equal to

(A) A ∩ (B ∪ C)
(B) A ∪ (B ∩ C)
(C) A ∩ (B – C)
(D) A – (B ∪ C)

2) Which of the following option(s) is/are true ?

(A) If A ⊂ B and A ≠ B, then A is called a proper subset of B

(B) If A ⊂ B, then B is called superset of A.
(C) A = {1, 2, 3} is NOT a proper subset of B = {1, 2, 3, 4}
(D) If a set A has only one element, we call it a singleton set.

3) If x2 – 3x + 2 is a factor of x4 – px2 + q, then-

(A) equation x4 – px2 + q = 0 has four distinct real roots

(B) equation x4 – px2 + q = 0 has two real and two imaginary roots
(C) p = – 5, q = – 4
(D) p = 5, q = 4

4) If one root of the equation 4x2 + 2x – 1=0 is ‘α’, then

α can be equal to
(A)
(B)
α can be equal to
(C) other root is 4α3 – 3α
(D) other root is 4α3 + 3α

SECTION-I (iii)

Common Content for Question No. 1 to 2

Let ƒ(x) = x2 + 3x + 1 and g(x) = x + 1.

1) Range of , x ∈ R – {–1} is -

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

2) Let ƒ(x) + λg(x) > –10 ∀ x ∈ R, then sum of all possible integral values of λ is -

(A) –11
(B) –13
(C) 11
(D) 13

Common Content for Question No. 3 to 4

Let ƒ(x) = 4x2 – 2x + k, where k ∈ R

3) If both roots of the equation ƒ(x) = 0 are negative, then k is -

(A)

(B)

(C) (–∞,0)
(D) ϕ

4) If both roots of equation ƒ(x) = 0 lie in (–1,1), then number of integers in the range of k are -

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

SECTION-III
1) Number of real solution(s) of equation (2x – 1)x2 +

2) If (x2 + 2x + 4)(y2 + 10y + 27) > k, ∀ x, y ∈ R, then maximum possible value of k is

3) If α,β are the roots of quadratic equation x2 – 4x + 1 = 0 such that an = αn + βn, then value of

4) If least common multiple of quadratic expressions x2 – ax + b and x2 – cx + 2a is x3 – 6x2 + 11x – 6

and greatest common divisor is (x – 2), then a + b – c is equal to

5) Number of integral value of x, for which equation |x2 – 4x + 3| + |x2 – 6x + 5| = 2|x – 1| satisfies, is

6) Number of integral values of x simultaneously satisfying the system of inequalities

and is

SECTION-IV

1) Let ƒ(x) = x2 + 5x + 1 and α, β are roots of equation ƒ(x) = 0, then

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

Column-I Column-II
Graph of ƒ(x) is symmetric with respect to
(A) (P) 7
x = k, then value of |2k| is
Let graph of ƒ(x) intersect x-axis at P and Q,
(B) (Q) 6
if length PQ is d then is equal to
(C) Minimum integral value of ƒ(x) is m then |m| is (R) 5
Number of integral values of x which satisfying
(D) (S) 4
ƒ(x) < 7 is
(T) 8

2) The values of k ∈ R for which Match Column-I with Column-II and select the correct answer using
the code given below the list.

Column-I Column-II

(A) sum of roots of x2 + (2 – k – k2)x – k2 = 0 is zero (P) –4

(B) The roots of x2 + (2k – 1)x + k2 + 2 = 0 are in ratio 1 : 2 (Q) 1

Sum of the square of the roots of x2 – (k – 2)x – k – 1 = 0

(C) (R)
assume the least value

(D) kx2 – (k + 1)x + 2k – 1 = 0 has equal roots (S) –2

(T) –3
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-I (iii)

Q. 9 10 11 12
A. C A B C

SECTION-III

Q. 13 14 15 16 17 18
A. 1 3 6 6 2 1

SECTION-IV

Q. 19 20
A. A->R,B->P,C->T,D->P A->P,B->S,C->Q,D->T

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 21 22 23 24
A. A C B A

SECTION-I (ii)

Q. 25 26 27 28
A. B,C A,B,D A,B C,D

SECTION-I (iii)

Q. 29 30 31 32
A. A C A A

SECTION-III

Q. 33 34 35 36 37 38
A. 1 7 7 5 4 5
 SECTION-IV

Q. 39 40
A. A->PQRS,B->QRT,C->QS,D->P A->Q,B->R,C->S,D->P

PART-3 : MATHEMATICS

SECTION-I (i)

Q. 41 42 43 44
A. C B D A

SECTION-I (ii)

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

SECTION-I (iii)

Q. 49 50 51 52
A. D A D B

SECTION-III

Q. 53 54 55 56 57 58
A. 3 6 4 0 4 1

SECTION-IV

Q. 59 60
A. A->R,B->P,C->R,D->Q A->QS,B->P,C->Q,D->QR
 SOLUTIONS

PART-1 : PHYSICS

1)

cos 1.8° = cos

=1–

1 – cos 1.8° =

2)

Check for dimensionally correct option.

5)

Velocity = length/time, acc = length/(time)2

and

Now,

Time = vel./acc, i.e., and

Momentum = mass × velocity, i.e.,

6)

For any physical quantity; numerical value × unit = constant

For (A) n1u1 = n2u2
⇒ (500)

= = 0.5

For (B)

For (C)

=
= 3.6 × 10–4

For (D)

= = 0.36

7)

(A)

(B) if

2α < θ

And β <
2β + θ < 180°

(C) if

2α > θ and β >

2β + θ > 180°

(D) if

and β =
α + β = 90°
 12) 90 LT–1 = xLgaj T–1kaal

15)

19) For (A) : Q[Acceleration] = LT–2 ∴ unit of acceleration = (1km) (1min)–2 = m/s2
For (B) : Q[Kinetic energy] = ML2T–2

∴ unit of kinetic energy = (1 Quintal) (1km)2(1min)–2 = × 105 kgm2s–2

For (C) : Q[Pressure] = ML–1T–2

∴ unit of pressure = (1 Quintal) (1km)–1(1min)–2 = × 10–4 kgm–1s–2

For (D) : Q[Work] = ML2T–2

∴ unit of work = (1 Quintal) (1km)2(1min)–2 = × 105 kgm2s–2

PART-2 : CHEMISTRY

21) Level : Easy

Concept :
Heavy water is D2O. Its molar mass = 16 + 2 2 = 20g
Normal water is H2O. Its molar mass = 18g
Solution :
Molar mass of heavy water = 20g
Molar mass of H2O = 18g
Difference in molar mass = 2g

Percentage different = = 10
Thus, molar mass of water is 10% less as compured to heavy water.
Hence, correct answer is option (A)

22)

initial molar =

moles added =

23)

Ca → = =1
Br → =1
⇒ CaBr2

24) Vgases = 400ml

25) For option (A)

Let atomic mass of P be MP and atomic mass of Q be MQ
Molar ratio of atoms P : Q in compound 3 is

Molar ratio of atoms P : Q in compound 2 is

= 44.4 MQ : 55.6 MP

= 44.4 MQ : 55.6 ×

= 44.4 : 55.6 ×
= 9 : 10
=> Empirical formula of compound 2 is therefore P9Q10
Option (A) in incorrect

For option (B)

Molar Ratio of atoms P : Q in compound 3 is

If MP = 20
Option (B) is correct

For option (C)

Molar ratio of atoms P : Q in compound 2 is

Molar ratio of atoms P : Q in compound 1 is

= 55.6 : 44.4
5:4
Hence, empirical formula of compound 1 is P5Q4
Hence, option (C) is correct
For option (D)
Molar ratio of atoms P : Q in compound 1 is

= 35 : 70 = 1 : 2
Hence, empirical formula of compound 1 is PQ2
Hence, option (D) is incorrect

26)

(A) Molarity of second solution is = 1 M

Molarity of final solution =

(B) Volume = 100 + 100 = 200 mL

(D) Mass of H2SO4 = × 98 = 19.6 g.

27)

C3H8+O2 → 100 ml
Case-I :-
Let assume C3H8 is L.R.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (l)
initial V1 V2
final 0 V2 – 5V1 3V1
Vol. contraction = Vi–Vf
VC = (V1+V2)–(V2–5V1+3V1)
45 = 3V1
V1 = 15 ml (C3H8)
V2 = 85 ml (O2)
Case-II :-
Let assume O2 is L.R.
C3H8 + 5O2 → 3CO2 + 4H2O (ℓ)
Initial V1 V2

Final V1 – 0 V2
Vol. contraction = Vi + Vf

45 =
V2 = 75 ml
V1 = 25 ml

28)

1 g atom of nitrogen means 1 mole atoms of nitrogen or NA atoms of nitrogen

(A) 6.02 × 1023 N2 molecule = 2 × 6.02 × 1023 atoms of nitrogen

(B) Moles = 1 mole N2 = 6.02 × 1023 N2 molecule =2 × 6.02 × 1023 atoms of nitrogen

(C) Moles = 0.5 moles N2 = 3.01 × 1023 N2 molecule =2 × 3.01 × 1023 atoms of
nitrogen = 6.02 × 1023 atoms of nitrogen

(D) Moles of N2 = = 0.5 moles N2 = 3.01 × 1023 N2 molecule =2 × 3.01 × 1023 atoms of
nitrogen = 6.02 × 1023 atoms of nitrogen

29)

CxHyOz

1y = 2z ... (1)
12x + y + 16z = 180 ... (2)
% C = 40 %

from (1) & (2)

z=6
y = 12
So, compound is
C6H12O6
Empirical formula
% by mass of oxygen

= 53.33 %

30)

CxHyOz

1y = 2z ... (1)
12x + y + 16z = 180 ... (2)
% C = 40 %

from (1) & (2)

z=6
y = 12
So, compound is
C6H12O6
Empirical formula

31) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2

 0.3 mol

= 3.36

32) =

33) O3 + 2KI + H2O → I2 + 2KOH + O2

intial 5 mole 6 mole 4 mole
After reaction 2 mole 1 mole 3 mole 6 mole 3 mole
3×22.4L

34) Vol. of silver atom =

=
Vol of Ag = 1.708 × 10–23 cm3

= 1.708 × 10–23

= 1.6 × 10–8 cm
Circular area of Ag atom = πr2 = π(1.6 × 10–8)2
= 8.04 × 10–20m2

Number of Ag atom in the surface =

y × 10x =
y × 10x = 1.24 × 107

35) Let assume 34Cl has x gram and 38Cl has

(7– x) gram
moles of 34Cl = x/34 and moles of
38
Cl = (7 – x) / 38

Mavg. =
put the value of Mavg. find the value of x,
so x = 5.1 g
moles of 34Cl = 5.1/34 = 0.15
and moles of 38Cl = (7 – 5.1)/38 =.05
n = 0.15 × 17 + 0.05 × 21 = 3.6
p = 0.15 × 17 + 0.05 × 17 = 3.4
∴ p+n=7
36)

0.06 =
x=5

37)

NH4OH .

38) C2H2 + → 2CO2 + H2O

x 2x
C2H4 + 3O2 → 2CO2 + 2H2O
20 – x 3(20 – X) 2(20 – x)
Vi – Vf = 37.5

x = 5 ml

39) (A) 120 g CH3COOH in 1 L solution

(P) M =
(Q) 1.2 × 1000 g solution contain → 120 g CH3COOH

100 g contain → = = 10%

(R) 1000 mL contain → 120 g

100 mL solution → × 100 = 12% (w/v)

(S) m = ⇒ (A) → P, Q, R, S
(B) 120 g Glucose (C6H12O6) in 1 lit. (d = 1.2)
From (A) It is clear that molarity and molality of glucose will be different because mole are
different but % (w/w) & % (w/v) will remain same.
∴ Density is same , B → (Q,R,T)

(C) xu =

(S) xu = or

⇒ m = 1.85
 (Q) wt. of solution = wt. of water + wt. of urea = 30 × 18 + 1 × 60 = 600
600 g solution contain urea = 60 g

600 g solution contain urea = = 10% (w/v)

(D) 19.6% (w/v) H2SO4 solution (d = 1.2 g/mL)
100 mL solution contain = 19.6 g H2SO4
1000 mL solution contain = 196 g

1 L solution contain = mol H2SO4 = 2 mole

40) (P) Mass of one molecule of H2SO4 = 98 amu.

(Q) Molar mass of H2SO4 = 98 gm
(R) 1 amu = 1.66 × 10–27 kg.
= 1.66 × 10–24 gm.
(S) Relative atomic mass of Mg = 24.

PART-3 : MATHEMATICS

41)
second equation becomes
2x2 + 10x + 24 = 0
D < 0 ⇒ imaginary roots

42) α.α3 = q ⇒
⇒
⇒ q1/2 + q3/2 + 2q = p2
3 2 2 4 2
⇒ q + q + 2q = 4q + p – 4p q
⇒ q3 – 2q2 – p4 + 4p2q + q = 0

43) It is obvious from figure that roots are of opposite sign but –ve root has greater magnitude

44) As the roots are imaginary, so both the roots are common

⇒
45)
A = (1) + (2) + (3) + (4)
B ∪ C = (2) + (3) + (4) + (5) + (6) + (7)
A – (B ∪ C) = {(1), (2), (3), (4)} – {(2), (3), (4), (5), (6), (7)}
= {(1)}

46) A ⊂ B and A ≠ B

A is called proper subset of B

B is called superset of A
A = {1, 2, 3}
B = {1, 2, 3, 4}
A ⊂ B ⇒ A is called proper subset of B
If set A has only are element, then we call it a singleton set.
Option A, B, D are correct

47) (x – 1) (x – 2) is a factor of x4 – px2 + q

⇒ x = 1, 2 are roots
⇒ 1 – p + q = 0 and 16 – 4p + q = 0
on solving, p = 5, q = 4
∴ equation : x4 – 5x2 + 4 = 0
⇒ (x2 – 4) (x2 –1) = 0
4 real and distinct roots.

48) 4x2 + 2x – 1 = 0

x=

If α = ⇒ α3 =

4α3 =
 ⇒ 4α3 – 3α =

49)
x2 + (3 – y)x + 1 – y = 0, x ∈ R
D>0
9 + y2 – 6y – 4 + 4y > 0
y2 – 2y + 5 > 0
∵ D < 0 ⇒ y ∈ R.

50)

x2 + (3 + λ)x + λ + 11 > 0 ∀ x ∈ R
∴ D<0
9 + λ2 + 6λ – 4λ – 44 < 0
(λ + 7)(λ – 5) < 0

Sum of integral values = –11

51) k ∈ ϕ

52)
2 integer

53) Given equation is

put x2 –1 = t
∴ (2x – 1)t + 2x(2t – 1) = 0
it is possible only when x = 0 or t = 0
∴ x = 1, –1, 0.

54)

[(x + 1)2 + 3][(y + 5)2 + 2] ≥ k

∴ k ≤ 6 when x + 1 = 0 and y + 5 = 0
∴ |k| = 6

55) α + β = 4, αβ=1

Let
 =α+β=4

56) L.C.M = (x – 1) (x – 2) (x – 3) given

H.C.F. = (x – 2) given
so the quadratic expressions are
(x – 1) (x – 2) & (x – 2) (x – 3)
⇒ x2 – 3x + 2 & x2 – 5x + 6
⇒ x2 – ax + b & x2 – cx + 2a
⇒ a = 3, b = 2, c = 5

57) (x2 – 4x + 3) (x2 – 6x + 5) < 0

⇒ (x – 1)2(x – 3) (x – 5) < 0
⇒ x ∈ [3,5] ∪ {1}

58)
......(1)

⇒
x ∈ (–∞, 1) ..........(2)
Taking intersection of (1) and (2)

x ∈ (–1, 1)

59) (A) ƒ(x) is symmetric with respect to ∴ 2k = –5

(B)
∴

(C) Range
∴ m = –5
(D) ƒ(x) < 7
x2 + 5x – 6 < 0
(x + 6)(x – 1) < 0
∴ x ∈ (–6, 1)

60) (A) 2 – k – k2 = 0 ⇒ k = 1, – 2
(B) α + 2α = –(2k – 1)
α(2α) = k2+ 2
(C) α2 + β2 = (k – 2)2 + 2(k +1)
= k2 – 2k + 6 = (k – 1)2 + 5
so for k= 1 it assume least value
(D) (k + 1)2 – 4k(2k – 1) = 0
7k2 – 6k – 1 = 0 ⇒ (7k + 1) (k – 1) = 0