[tabs title=”JEE MAINS Previous Year Paper” type=”centered”]

[tab title=”PHYSICS”]

PHYSICS

SECTION A

Q 1: A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use g = GM/R2 = 9.8 ms-2 and radius of earth, R = 6400 km.]

(A) 49 N

(B) 49.83 N

(C) 49.17 N

(D) 48.83 N

Answer: (D)
At North Pole, weight
Mg = 49
Now, at equator
g’ = g – ω²R
⇒ Mg’ = M(g – ω²R)
⇒ weight will be less than Mg at equator.
Alter:
g is maximum at the poles.
Hence from options only (D) has lesser value than 49N.

Q 2: The logic circuit shown below is equivalent to :

Answer: (2)

Q 3: If one mole of an ideal gas at (P₁, V₁) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B→C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to:

(A) 0

(B) -RT/2( γ – 1)

(C) RT [ln2 -1/2( γ -1)]

(D) RT ln2

Answer: (C)
AB → Isothermal process
W_AB = nRT ln2 = RT ln2
BC → Isochoric process
W_BC = 0
CA → Adiabatic process
W_CA = P₁V₁ – (P₁/4)X²V₁)/(1 – γ)
= P₁V₁/2(1 – γ)
= RT/2(1 – γ )
W_ABCA = RT ln2 + RT/2(1 – γ)
= RT [ln2 – 1/2(γ – 1)]

Q 4: Match List – I with List – II.

List – I	List – II
(a) Source of microwave frequency	(i) Radioactive decay of nucleus
(b) Source of infrared frequency	(ii) Magnetron
(c) Source of Gamma Rays	(iii) Inner shell electrons
(d) Source of X-rays	(iv) Vibration of atoms and molecules
–	(v) LASER
–	(vi) RC circuit

Choose the correct answer from the options given below:

(A) (a)-(ii),(b)-(iv),(c)-(i),(d)-(iii)

(B) (a)-(vi),(b)-(iv),(c)-(i),(d)-(v)

(C) (a)-(ii),(b)-(iv),(c)-(vi),(d)-(iii)

(D) (a)-(vi),(b)-(v),(c)-(i),(d)-(iv)

Answer: (A)
(a) Source of microwave frequency – (ii) Magnetron
(b) Source of infrared frequency – (iv) Vibration of atom and molecules
(c) Source of gamma ray – (i) Radioactive decay of nucleus
(d) Source of X-ray – (iii) inner shell electron

Q 5: The period of oscillation of a simple pendulum is T = 2π√(L/g). Measured value of ‘L’ is 1.0 m from meter-scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of ‘g’ will be:

(A) 1.33 %

(B) 1.30 %

(C) 1.13 %

(D) 1.03 %

Answer: (C)
T = 2π√(L/g)
T² = 4𝜋² [L/g]
g = 4𝜋² [L/T2]
Δg/g = ΔL/L + 2ΔT/T
= [1mm/1m + 2(10×10-3)/1.95]×100
= 1.13 %

Q 6: When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:

(A) elliptical

(B) parabolic

(C) straight line

(D) circular

Answer: (A)
We know that in SHM;
V = ω√(A² – x²)

Elliptical
Alternate:
x = A sinωt ⇒ sin ωt = x/A
v = Aω cos ωt ⇒ cos ωt = v/Aω
Hence (x/A)² + (v/Aω)² = 1
Which is the equation of an ellipse.

Q 7: In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

(A) (1/2π) √(2K/Mg sin α)

(B) (1/2π) √(K/Mg sin α)

(C) (1/2π) √(2K/M)

(D) (1/2π) √(K/2M)

Answer: (A)
Equivalent K = K + K = 2K
Now, T = 2π √(M/K_eq)
⇒ T = 2π √(M/2K)
∴f = (1/2π) √(2K/M)

Q 8: Given below are two statements:

Statement I: PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal.

Statement II: In the study of transistor, the amplification factor β indicates ratio of the collector current to the base current.

In the light of the above statements, choose the correct answer from the options given below.

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

(B) Both Statement I and Statement II are true.

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

(D) Both Statement I and Statement II are false.

Answer: (A)
S-1
Statement 1 is false because in case of two discrete back to back connected diodes, there are four doped regions instead of three and there is nothing that resembles a thin base region between an emitter and a collector.
S-2
Statement-2 is true, as
β = I_C/I_B

Q 9: Figure shows a circuit that contains four identical resistors with resistance R = 2.0 Ω. Two identical inductors with inductance L = 2.0 mH and an ideal battery with emf E = 9.V. The current ‘i’ just after the switch ‘s’ is closed will be:

(A) 9A

(B) 3.0 A

(C) 2.25 A

(D) 3.37 A

Answer: (C)
Just when switch S is closed, inductor will behave like an infinite resistance. Hence, the circuit will be like

Given: V = 9 V
From V = IR
I = V/R
R_eq. = 2 + 2 = 4 Ω
i = 9/4 = 2.25 A

Q 10: A circular hole of radius (a/2) is cut out of a circular disc of radius ‘a’ shown in figure. The centroid of the remaining circular portion with respect to point ‘O’ will be:

(A) (10/11)a

(B) (⅔)a

(C) (⅙)a

(D) (⅚)a

Answer: (D)
Let σ be the surface mass density of disc.

Xcom= (m₁x₁ – m₂x₂)/(m₁ – m₂)
Where m = σπr²
Xcom= (σ×πa²×a) – (σπa²/4) × 3a/2)/(σπa² – σπa²/4)
Xcom= (a – 3a/8)/(1 – ¼)
Xcom= (5a/8)/(¾)
Xcom= 5a/6

Q 11: The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is:

(A) 4:2

(B) 4:1

(C) 1:4

(D) 4:3

Answer: (B)
From De-Broglie’s wavelength:-
λ = h/mv
Given λP = λα
v ∝ 1/m
v_p/v_α = m_α/m_p = 4m_p/m_p
= 4/1

Q 12: On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

(A) suffer change in momentum when impinge on the walls of container.

(B) continuously stick to the walls of container.

(C) continuously lose their energy till it reaches wall.

(D) are attracted by the walls of container.

Answer: (A)
Based on kinetic theory of gases, molecules suffer change in momentum when impinge on the walls of container. Due to this they exert a force resulting in exerting pressure on the walls of the container.

Q 13: Two electrons each are fixed at a distance ‘2d’. A third charge proton placed at the midpoint is displaced slightly by a distance x (x<<d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:

(m = mass of charged particle)

(A) (q²/2πε₀md³)^1/2

(B) (πε₀md³/2q²)^1/2

(C) (2πε₀md³/q²)^1/2

(D) (2q²/πε₀md³)^1/2

Answer: (A)

Restoring force on proton:-
Fr = 2F₁ sinθ where F₁ = kq²/(d² + x²)
Fr = 2Kq²x/[d² + x²]^3/2
x <<< d
Fr = 2kq²x/d³
= q²x/2πε₀d³
= kx
K = q²/2πε₀d³
Angular Frequency:-
ω = √(k/m)
ω = √(q²/2πε₀md³)

Q 14: An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:

(A) 10^-2 nm

(B) 10^-3 nm

(C) 10^-4 nm

(D) 10^-1 nm

Answer: (B)

The minimum wavelength of photon will correspond to the maximum energy due to accelerating by V volts in the tube.
λmin = hc/eV
λmin = 1240nm-eV/1.24×10⁶
λmin = 10^-3 nm

Q 15: A soft ferromagnetic material is placed in an external magnetic fiel(D) The magnetic domains:

(A) decrease in size and changes orientation.

(B) may increase or decrease in size and change its orientation.

(C) increase in size but no change in orientation.

(D) have no relation with external magnetic field.

Answer: (B)
Atoms of ferromagnetic material in unmagnetized state form domains inside the ferromagnetic material. These domains have large magnetic moment of atoms. In the absence of magnetic field, these domains have magnetic moment in different directions. But when the magnetic field is applied, domains aligned in the direction of the field grow in size and those aligned in the direction opposite to the field reduce in size and also its orientation changes.

Q 16: According to Bohr atom model, in which of the following transitions will the frequency be maximum?

(A) n = 2 to n = 1

(B) n = 4 to n = 3

(C) n = 5 to n = 4

(D) n = 3 to n = 2

Answer: (A)

Since, ΔE is maximum for the transition from n = 2 to n = 1
f is more for transition from n = 2 to n = 1.

Q 17: Which of the following equations represents a travelling wave?

(A) y = Ae^{x^2} (vt + θ )

(B) y = A sin(15x – 2t)

(C) y = Aex cos(ωt – θ)

(D) y = A sin x cos ωt

Answer: (B)
Y = F(x, t)
For travelling wave y should be linear function of x and t and they must exist as (x ± vt)
Y = A sin (15x – 2t) which is a linear function in x and t.

Q 18: Zener breakdown occurs in a p-n junction having p and n both:

(A) lightly doped and have wide depletion layer.

(B) heavily doped and have narrow depletion layer.

(C) heavily doped and have wide depletion layer.

(D) lightly doped and have narrow depletion layer.

Answer: (B)
The zener breakdown occurs in the heavily doped p-n junction diode. Heavily doped p-n junction diodes have narrow depletion region. The narrow depletion layer width leads to a high electric field which causes the p-n junction breakdown.

Q 19: A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = -αx². The distance at which the particle stops:

(A) (2v₀/3α)^1/3

(B) (3v₀²/2α)^1/2

(C) (3v₀²/2α)^1/3

(D) (2v₀²/3α)^1/2

Answer: Bonus
a = vdv/dx

Given:- v_i = v₀
V_f = 0
X_i = 0
X_f = x
From Damping Force: a = -αx²/m

-v₀²/2 = (-α/m) [x³/3]
x = [3mv₀²/2α]^1/3
Most suitable answer could be (3) as mass ‘m’ is not given in any options.

Q 20: If the source of light used in a Young’s double slit experiment is changed from red to violet:

(A) the fringes will become brighter.

(B) consecutive fringe lines will come closer.

(C) the central bright fringe will become a dark fringe.

(D) the intensity of minima will increase.

Answer: (B)
β = λD/d
As λv<λR
βv< βR
⇒ Consecutive fringe lines will come closer.
⇒ (2)

Section – B

Q 1: A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is ______×10^-1 kg m².

Answer: 8

MOI of AB about P : IABP = (M/6)(l/6)²/12
MOI of AB about O,

= (6/100) [(24×24/12×36) + (24×24/36)×3/4]
= 0.8 kg m²
= 8×10^-1 kg m²

Q 2: Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be A/1. So the value of A will be ________.

Answer: 2
Given that, M₁/M₂ = 1/2
we know that
K = p²/2M
K₁/K₂ = (p²/2M₁ )×2M₂/p²
K₁/K₂ = M₂/M₁ = 2/1
A/1 = 2/1
⇒∴ A = 2

Q 3: An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _______ ×10^-2 cm.

Answer: 667
ε_r = 2.25
Assuming non-magnetic material ⇒ μ_r = 1
Hence refractive index of the medium
n = √(μ_rε_r )= √2.25 = 1.5
∴λ_v/λ_m=n
λ_m = C/f.n
= 3×10⁸/3×10⁹×1.5
= (⅔) 10^-1 m
= 667×10^-2 cm

Q 4: The root mean square speed of molecules of a given mass of a gas at 270C and 1 atmosphere pressure is 200 ms-1. The root mean square speed of molecules of the gas at 1270C and 2 atmosphere pressure is x/√3 ms-1. The value of x will be __________.

Answer: 400 m/s
Vrms = √(3RT₁/M₀)
200 = √(3R×300/M₀) ….(1)
Also, x/√3 = √(3R×400/M₀) …(2)
(1)÷(2)
200/(x/√3) = √(300/400) = √(3/4)
⇒x = 400 m/s

Q 5: A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5 dB per km and cable length is 20 km. The power received at the receiver is 10^-xW. The value of x is ______.

[Gain in dB = 10 log₁₀(P₀/P_i)]

Answer: 8
Power of signal transmitted: P_i = 0.1 Kw = 100w
Rate of attenuation = -5 dB/Km
Total length of path = 20 km
Total loss suffered = -5×20 = -100dB
Gain in dB = 10 log₁₀(P₀/P_i)
-100 = 10log₁₀(P₀/P_i)
log₁₀(P_i/P₀) = 10
log₁₀(P_i/P₀) = log101010
100/P₀ = 1010
⇒ P₀ = 1/10⁸ = 10^-8
⇒ x = 8

Q 6: A series LCR circuit is designed to resonate at an angular frequency ω0 = 10⁵rad/s. The circuit draws 16W power from 120 V source at resonance. The value of resistance ‘R’ in the circuit is _______ Ω.

Answer: 900
P = V²/R
16 = 120²/R
⇒ R = 14400/16
⇒ R = 900 Ω

Q 7: A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force, will be ________ cm.

Answer: 2

y = Fl/A∆l
F/A = y∆l/l
F/A = y×0.04/l …(1)
When length & diameter is doubled.
⇒ F/4A = y × ∆l/2l …(2)
(1)÷(2)
(F/A)/(F/4A) = (y×0.04/l)y×∆l/2l
4 = 0.04×2/∆l
∆l = 0.02
∆l = 2×10^-2
∴ x = 2

Q 8: A cylindrical wire of radius 0.5 mm and conductivity 5×107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3π mA. The value of x is ____.

Answer: 5
We know that
J = σE
⇒ J = 5×10⁷×10×10^-3
⇒ J = 50×104 A/m²
Current flowing;
I = J × πR²
I = 50×10⁴ ×π(0.5×10^-3)²
I = 50×10⁴×π×0.25×10^-6
I = 125×10^-3π
X = 5

Q 9: A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be _____ ×10³ Nm²/C.

Answer: 226
Using Gauss law, it is a part of cube of side 12 cm and charge at centre,
ϕ = Q/6ε₀= 12μc/6ε₀

12×10⁻⁶ /6×8.85×10⁻¹²​

= 226×103 Nm²/C

Q 10: Two cars are approaching each other at an equal speed of 7.2 km/hr. When they see each other, both blow horns having frequency of 676 Hz. The beat frequency heard by each driver will be ________ Hz. [Velocity of sound in air is 340 m/s.]

Answer: 8

Speed = 7.2 km/h = 2 m/s
Frequency as heard by A
f’A= fB(v + v₀)/(v – v_s)
f’A = 676(340 + 2)/(340 – 2)
f’A = 684Hz
∴ fBeat = f’A- fB
= 684 – 676
= 8 Hz

[/tab]

[tab title=”CHEMISTRY”]

Chemistry

SECTION A

Q 1. Given below are two statements:

(A) Both Statement I and Statement II are false

(B) Statement I is false but Statement II is true

(C) Statement I is true but Statement II is false

(D) Both Statement I and Statement II are true

Answer: (C)
For survival of aquatic life dissolved oxygen is responsible for its optimum limit 6.5 ppm and optimum limit of BOD ranges from 10-20 ppm & BOD stands for biochemical oxygen demand.

Q 2. Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of HgSO₄ and H₂SO₄?

Answer: (A)
Reaction of Alkyne with HgSO₄ & H₂SO₄ follow as

Hence, by this process preparation of CH₃CH₂CHO can’t be possible.

Q 3. Which one of the following compounds is non-aromatic?

Answer: (B)

(Not planar)
Hence, it is non-aromatic.

Q 4. The incorrect statement among the following is:

(A) VOSO₄ is a reducing agent

(B) Red color of ruby is due to the presence of CO³⁺

(C) Cr₂O₃ is an amphoteric oxide

(D) RuO₄ is an oxidizing agent

Answer: (B)
Red color of ruby is due to presence of CrO₃ or Cr⁺⁶ not CO³⁺

Q 5. According to Bohr’s atomic theory:

(a) Kinetic energy of electron is ∝ Z² / n²

(b) The product of velocity (v) of electron and principal quantum number (n). ‘vn’ ∝ z²

(c) Frequency of revolution of electron in an orbit is ∝ Z³ / n³

(d) Coulombic force of attraction on the electron is ∝ Z³ / n⁴

Choose the most appropriate Answer from the options given below

(A) (C) only

(B) (A) and (D) only

(C) (A) only

(D) (A), (C) and (D) only

Answer: (B)
(a) KE = –TE = 13.6 × Z² / n² eV
KE α Z² / n²
(b) V = 2.188 × 10⁶ × z / n m/s
So, V_n ∝ Z
Frequency = V / 2πr
F α Z² / n² [∴ r α Z² / n² and v α Z / n ]
(d) Force ∝ Z² / r²
So, F ∝ Z³ / n⁴
So, only statement (A) is correct.

Q 6. Match List-I with List-II

List- I	List-II
(a) Valium	(iv) Tranquilizer
(b) Morphine	(iii) Analgesic
(c) Norethindrone	(i) Antifertility drug
(d) Vitamin B12	(ii) Pernicious anemia

(A) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)

(B) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)

(C) (A)-(ii), (B)-(iv), (C)-(iii), (D)-(i)

(D) (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)

Answer: (D)
(a) Valium (iv) Tranquilizer
(b) Morphine (iii) Analgesic
(c) Norethindrone (i) Antifertility drug
(d) Vitamin B₁₂ (ii) Pernicious anemia

Q 7. Match List-I with List-II

List- I (Salt)	List-II(Flame colour wavelength)
(a) LiCl	(i) 455.5 nm
(b) NaCl	(ii) 970.8 nm
(c) RbCl	(iii) 780.0 nm
(d) CsCl	(iv) 589.2 nm

Choose the correct Answer from the options given below:

(A) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(B) (A)-(ii), (B)-(iv), (C)-(iii), (D)-(i)

(C) (A)-(iv), (B)-(ii), (C)-(iii), (D)-(i)

(D) (A)-(i), (B)-(iv), (C)-(ii), (D)-(iii)

Answer: (B)
Range of visible region: 390 nm – 760 nm
VIBGYOR
Violet – Red
LiCl Crimson Red
NaCl Golden yellow
RbCl Violet
CsCl Blue
So, LiCl which is crimson have wavelengths close to red in the spectrum of visible region which is as per given data.

Q 8. Match List-I and List-II.

List-I	List-II
(A)	(i) Br2 / NaOH
(B)	(ii) H2 / Pd-BaSO4
(C)	(iii) Zn (Hg) / Conc. HCl
(D)	(iv) Cl2 / Red P, H2O

(A) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(B) (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)

(C) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Answer: (D)

Q 9. In polymer Buna-S: ‘S’ stands for

(A) Styrene

(B) Sulphur

(C) Strength

(D) Sulphonation

Answer: (A)
Buna-S is the co-polymer of buta-1,3-diene & styrene

Q 10. Most suitable salt which can be used for efficient clotting of blood will be:

(A) Mg(HCO₃)₂

(B) FeSO₄

(C) NaHCO₃

(D) FeCl₃

Answer: (D)
Blood is a negative sol, according to Hardy-Schulz’s rule, the cation with high charge has high coagulation power. Hence, FeCl₃ can be used for clotting blood.

Q 11. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

Reason R: Hydrogen is the lightest element.

In the light of the above statements, choose the correct Answer from the given below

(1) A is false but R is true

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

(3) A is true but R is false

(4) Both A and R are true but R is NOT the correct explanation of A

(A) A is false but R is true

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

(C) A is true but R is false

(D) Both A and R are true but R is NOT the correct explanation of A

Answer: (B)
Hydrogen is the most abundant element in the universe because all luminous bodies of the universe i.e. stars and nebulae are made up of hydrogen which acts as nuclear fuel and fusion reaction is responsible for their light.

Q 12. What is the correct sequence of reagents used for converting nitrobenzene into m- dibromobenzene?

Answer: (D)

Q 13. The correct shape and I-I-I bond angles respectively in I^–₃ ion are:

(A) Trigonal planar; 120^o

(B) Distorted trigonal planar; 135^o and 90^o

(C) Linear; 180º

(D) T-shaped; 180º and 90º

Answer: (C)
I^–₃ has sp³d hybridization (2 BP + 3 LP) and linear geometry.

Q 14. What is the correct order of the following elements with respect to their density?

(A) Cr < Fe < Co < Cu < Zn

(B) Cr < Zn < Co < Cu < Fe

(C) Zn < Cu < Co < Fe < Cr

(D) Zn < Cr < Fe < Co < Cu

Answer: (D)
Fact Based
Density depends on many factors like atomic mass. atomic radius and packing efficiency.

Q 15. The Correct set from the following in which both pairs are in correct order of melting point is

(A) LiF > LiCl ; NaCl > MgO

(B) LiF > LiCl ; MgO > NaCl

(C) LiCl > LiF ; NaCl > MgO

(D) LiCl > LiF ; MgO > NaCl

Answer: (B)
Generally,
M.P. ∝ Lattice energy = KQ₁Q₂ / r⁺ + r^–​

∝ (packing efficiency)

Q 16. The calculated magnetic moments (spin only value) for species

[FeCl₄]^2–, [Co(C₂O₄)³]^3– and MnO_2–⁴ respectively are:

Answer: (C)

[FeCl₄]^2–
Fe²+ 3d⁶ → 4 unpaired electrons. as Cl^– in a weak field liquid.

Q 17. Which of the following reagent is suitable for the preparation of the product in the following reaction?

(A) Red P + Cl₂

(B) NH₂-NH₂/ C₂H₅O^–Na⁺

(C) Ni/H₂

(D) NaBH₄

Answer: (B)

It is a wolf-kishner reduction of carbonyl compounds.

Q 18. The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH?

Answer: (C)

Orange bright dye.

Q 19. The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is:

(A) (iv) < (i) < (iii) < (ii)

(B) (iv) < (i) < (ii) < (iii)

(C) (i) < (ii) < (iii) < (iv)

(D) (iv) < (iii) < (ii) < (i)

Answer: (C)

Reactivity ∝ – M group present at o/p position.

Q 20. Match List-I with List-II

List- I(Metal)	List-II(Ores)
(a) Aluminum	(i) Siderite
(b) Iron	(ii) Calamine
(c) Copper	(iii) Kaolinite
(d) Zinc	(iv) Malachite

(A) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)

(B) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)

(C) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Answer: (C)
Siderite FeCO₃
Calamine ZnCO₃
Kaolinite Si₂Al₂O₅(OH)₄ or Al₂O₃.2SiO₂.2H₂O
Malachite CuCO₃.Cu(OH)₂

Section B

Q 1. The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O₂ for complete oxidation and produces 4 times its own volume of CO₂ is C_xH_y. The value of y is

Answer: 8
C_xH_y + 6O₂ 4CO₂ + y/2 H₂O
Applying POAC on ‘O’ atoms
6 × 2 = 4 × 2 + y/2 × 1
y/2 = 4 ⇒ y = 8

Q 2. The volume occupied by 4.75 g of acetylene gas at 50°C and 740 mmHg pressure is _______L. (Rounded off to the nearest integer)

(Given R = 0.0826 L atm K^–1 mol^–1)

Answer: 5
T = 50C = 323.15 K
P = 740 mm of Hg = 740 / 760 atm
V = ?
moles (n) = 4.75 / 26 atm
V = 4.75 / 26 × 0.0821 × 323.15 / 740 × 760
V = 4.97 5 Lit

Q 3. Sucrose hydrolysis in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25ºC. After 9h, the fraction of sucrose remaining is f. The value of is _________× 10^–2 (Rounded off to the nearest integer)

[Assume: ln10 = 2.303, ln2 = 0.693]

Answer: 81

Sucrose  ^__(Hydrolysis)→  Glucose + Fructose
t_1/2 = 3.33h = 10 / 3h

C_t = C₀ / {2 × T/t_1/2}

Fraction of sucrose remaining = f = Ct/Co = 1/ {2 × T/t_1/2}

1/f = 2^(t/t_1/2)

Log(1/f) = log(2^(t/t_1/2)) = t/t_1/2 log(2)

9/ (10/3) × 0.3 – 8.1/10 – 0.81

= x × 10^–2

x = 81

Q 4. The total number of amines among the following which can be synthesized by Gabriel synthesis is _______

Answer: 3
Only 1^o amines can be prepared by Gabriel synthesis.

Q 5. 1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is ____× 10^–2.

Answer: 243

93 g Aniline produce 135 g acetanilide
1.86 g produce 135 × 1.86 / 93 = 2.70 g
At 10% loss, 90% product will be formed after purification.

Q 6. Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is ______.

(a) α-sulphur (B) β-sulphur (C) S₂-form

Answer: (A)
S₂ is like O₂ i.e. paramagnetic as per molecular orbital theory.

Q 7. C₆H₆ freezes at 5.5ºC. The temperature at which a solution of 10 g of C₄H₁₀ in 200 g of C₆H₆ freeze is ________ C. (The molal freezing point depression constant of C₆H₆ is 5.12C/m).

Answer: 1
ΔT_f = i × K_f × m
= 1 × 5.12 × 10 / 58200 × 1000
∆T_f = 5.12 × 50 / 58 = 4.414
T_f(solution) = T_k(solvent) – ΔT = 5.5 – 4.414 = 1.086oC
≈ 1.09°C = 1 (nearest integer)

Q 8. Assuming ideal behavior, the magnitude of log K for the following reaction at 25ºC is x × 10^–1. The value of x is __________. (Integer Answer)

3HC ≡ CH(g) ⇌ C₆H₆ (l)

[Given: Δ_fG° (HC = CH) = – 2.04 × 10⁵] mol^–1; 

Δ_fG°(C₆H₆) = – 1.24 × 105 J mol^–1;

R = 8.314 J K^–1 mol^–1] 

Answer: 855
ΔG°_r = ΔG°_f [C₆H₆ (l)] – 3 × ΔG°_f [HC = CH]

Q 9. The magnitude of the change in oxidising power of the MnO^4-/ Mn²⁺ couple is x × 10^–4 V, if the H+ concentration is decreased from 1M to 10^–4 M at 25°C. (Assume concentration of MnO^4- and Mn²⁺ to be same on change in H⁺ concentration). The value of x is _____. (Rounded off to the nearest integer).

[Given: 230RT / F = 0.059]

Answer: 3776
5e- + MnO^–₄ + 8H → Mn⁺² + 4H₂O

= 3776 × 10^–4
So, x = 3776

Q 10. The solubility product of PbI₂ is 8.0 × 10^–9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is ________ (Rounded off to the nearest integer)

Given: √2 = 1.41

Answer: 141
PbI₂(s) ⇌ Pb²⁺ (aq) + 2I^– (aq)
S+0.1 2s
K_SP (PbI₂) = 8 × 109
K_SP = [Pb⁺²][I^–]²
8 × 10^–9 = (S + 0.1) (2S)² ⇒ (8 × 10^–9 + 0.1) × 4S²
⇒ S² = 2 × 10^–8
S = 1.414 × 10^–4 mol/Lit
= x × 10^–6 mol/Lit
∴ x = 141.4 141

[/tab]

[tab title=”MATHS”]

Maths

SECTION A

Q 1: Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax² + bx + 1 = 0, then the value of α² + β² – αβ is:

(A) 71/256

(B) -69/256

(C) 69/256

(D) -71/256

Answer: (D)
2b = a + c
(2a + 2)/3 = 10/3 and (2b + c)/3 = 7/3
⇒ a = 4
2b + c = 7, 2b – c = 4 },solving
b = 11/4 and c = 3/2
∴ Quadratic equation is 4x²+ (11/4)x + 1 = 0
∴ The value of (α + β)² – 3αβ = (121/256) – (¾)
= -71/256

Q 2: The value of the integral, where [x] denotes the greatest integer less than or equal to x , is :

(A) –4

(B) –5

(C) -√2 – √3 – 1

(D) -√2 – √3 + 1

Answer: (C)

I =

Put x – 1 = t ; dx = dt

I = -6 + (√2 – 1) + 2√3 – 2√2 + 6 – 3√3
I = -1 – √2 – √3

Q 3: Let f: R → R be defined as

Let A = {x ∈ R ∶ f is increasing}. Then A is equal to :

(A) (-5, -4) ∪ (4,∞)

(B) (-5, ∞)

(C) (-∞, -5) ∪ (4, ∞)

(D) (-∞, -5) ∪ (-4, ∞)

Answer: (A)
f’(x) = {-55 ; x< -5 6(x² – x – 20) ; -5 < x < 4 6(x²– x – 6) ; x > 4
⇒ f’(x) = {-55 ; x< -5 6(x – 5)(x + 4) ; –5 < x < 4 6(x – 3)(x + 2) ; x > 4
Hence, f(x) is monotonically increasing in (-5, -4) ∪ (4, ∞)

Q 4: If the curve y = ax² + bx + c,x ∈ R passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:

(A) a = 1, b = 1, c = 0

(B) a = -1, b = 1, c = 1

(C) a = 1, b = 0, c = 1

(D) a = 1/2, b = 1/2 ,c = 1

Answer: (A)
2 = a + b + c
dy/dx = 2ax + b, (dy/dx)(0,0) = 1
⇒ b = 1 and a + c = 1
Since (0, 0) lies on curve,
∴ c = 0, a = 1
TRICK: (0, 0) lies on the curve. Only option (1) has c = 0

Q 5: The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is:

(A) 65/2⁷

(B) 135/2⁹

(C) 65/2⁸

(D) 35/2⁷

Answer: (B)
Let A and B be two subsets.
For each x∈{1,2,3,4,5} , there are four possibilities:
x ∈ A∩B, x∈ A’∩B, x ∈ A∩B’, x ∈ A’∩B’
So, the number of elements in sample space =45
Required probability
= (⁵C₂ × 3³)/4⁵
= (10×27)/2¹⁰
= 135/2⁹

Q 6: The vector equation of the plane passing through the intersection of the planes 

and

and the point (1, 0, 2) is:

Answer: (B)
Family of planes passing through the intersection of planes is

The above curve passes through

(3 − 1) + λ(1 + 2) = 0
⇒ λ = -2/3
Hence, equation of plane is

⇒

TRICK: Only option (2) satisfies the point (1, 0, 2)

Q 7: If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x − 1, then the coordinates of P are :

(A) (–2, 8)

(B) (1, 5)

(C) (3, 13)

(D) (2, 8)

Answer: (D)
Tangent at P is parallel to the given line.
dy/dx|P = 4
⇒2x₁ = 4

⇒x₁ = 2
Required point is (2, 8)

Q 8: Let A and B be 3×3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2 − B2A2)X = O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has:

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

Answer: (C)
AT = A, BT = –B
Let A²B² − B²A²= P
PT = (A²B² – B²A²)T
= (A²B²)T – (B²A²)T
= (B²)T (A²)T– (A²)T (B²)T
= B²A² – A²B
⇒ P is a skew-symmetric matrix.

∴ ay + bz = 0 …(1)
–ax + cz = 0 …(2)
–bx – cy =0 …(3)
From equation (1), (2), (3)
Δ = 0 and Δ₁ = Δ₂= Δ₃ = 0
∴ System of equations has infinite number of solutions.

Q 9: If n ≥ 2 is a positive integer, then the sum of the series n+¹C₂ + 2(²C₂ + ³C₂ + ⁴C₂ + …. + ⁿC₂) is

(A) n(n + 1)² (n + 2)/12

(B) n(n – 1)(2n + 1)/6

(C) n(n + 1)(2n + 1)/6

(D) n(2n + 1)(3n + 1)/6

Answer: (C)
²C₂ = ³C₃
Let S = ³C₃ + ³C₂ + ……. + ⁿC₂ = n+¹C₃ (∵ ⁿC_r + ⁿC_r–1 = ⁿ⁺¹C_r)
∴ ⁿ⁺¹C₂ + ⁿ⁺¹C₃ + ⁿ⁺¹C₃
= ⁿ⁺²C₂ + ⁿ⁺¹C₃
=(n + 2)!/3!(n – 1)! + (n + 1)!/3!(n – 2)!
=(n + 2)(n + 1)n/6 + (n + 1)(n)(n – 1)/6 = (n(n + 1)(2 + 1))/6
TRICK : Put n = 2 and verify the options.

Q 10: If a curve y = f(x) passes through the point (1, 2) and satisfies xdy/dx + y = bx4, then for what value of b,  

(A) 5

(B) 62/5

(C) 31/5

(D) 10

Answer: (D)
dy/dx + y/x = bx³
I.F. =

= x
∴ yx = ∫bx⁴ dx = bx⁵/5 + c
The above curve passes through (1, 2).
2 = b/5 + c
Also,

⇒ (b/25) ×32 + c ln 2 – b/25 = 62/5
⇒ c = 0 and b = 10

Q 11: The area of the region: R{(x, y): 5x² ≤ y ≤ 2x² + 9} is:

(A) 9√3 square units

(B) 12√3 square units

(C) 11√3 square units

(D) 6√3 square units

Answer: (B)

Required area
​

= 12√3

Q 12: Let f(x) be a differentiable function defined on [0,2] such that f’(x) =f ‘(2 – x) for all x∈(0, 2), f(0) = 1 and f(2) = e². Then the value of is:

(A) 1 + e²

(B) 1 – e²

(C) 2(1 – e²)

(D) 2(1+e²)

Answer: (A)
f'(x) = f'(2 – x)
On integrating both sides, we get
f(x) = -f(2 – x)+c
Put x = 0
f(0) + f(2) = c
⇒ c = 1+ e²
⇒ f(x) + f(2 – x) = 1 + e²
I =

= 1+e²

Q 13: The negation of the statement ~p∧(p∨q) is∶

(A) ~p∧q

(B) p∧~q

(C) ~p∨q

(D) p∨∼q

Answer: (D)
Negation of ~p∧(p∨q) is
∼[~p∧(p∨q)]
≡ p∨ ∼(p∨q)
≡ p∨(∼p∧∼q)
≡ (p∨∼p)∧(p∨∼q)
≡ T∧(p∨∼q), where T is tautology.
≡ p∨∼q

Q 14: For the system of linear equations: x − 2y = 1, x − y + kz = −2, ky + 4z = 6,k ∈ R

Consider the following statements:

(A) The system has unique solution if k ≠ 2,k ≠ −2.

(B) The system has unique solution if k = -2.

(C) The system has unique solution if k = 2.

(D) The system has no-solution if k = 2.

(E) The system has infinite number of solutions if k ≠ -2.

Which of the following statements are correct?

(A) (B) and (E) only

(B) (C) and (D) only

(C) (A) and (D) only

(D) (A) and (E) only

Answer: (C)
x -2y + 0.z = 1
x – y + kz = -2
0.x + ky + 4z = 6
 

= 4 – k2
For unique solution, 4 – k² ≠ 0
k ≠ ±2
For k = 2,
x – 2y + 0.z =1
x – y + 2z = -2
0.x + 2y + 4z = 6

= -8 + 2 (-20)
⇒Δx= -48 ≠ 0
For k = 2,Δx ≠ 0
So, for k = 2, the system has no solution.

Q 15: For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2)?

(A) x² + 9y² = 9

(B) 2x² – 18y² = 9

(C) y² = x/(6√3)

(D) x² + y² = 7

Answer: (A)
Tangent to x² + 9y² = 9 at point (3√3/2, 1/2) is x(3√3)/2 + 9y(1/2) = 9
⇒3√3 x + 9y = 18
⇒x +√3 y = 2√3
⇒ Option (A) is true.

Q 16: The angle of elevation of a jet plane from a point A on the ground is 600. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 300. If the jet plane is flying at a constant height, then its height is:

(A) 1200√3 m

(B) 1800√3 m

(C) 3600√3 m

(D) 2400√3 m

Answer: (A)

v = 432 × 1000/(60×60) m/sec
= 120 m/sec
Distance PQ = v × 20 = 2400 m
In ΔPAC
tan 60⁰ = h/AC
⇒ AC = h/(√3)
In ΔAQD
tan 30⁰ = h/AD
⇒AD = √3h
AD = AC + CD
⇒√3 h = h/√3 + 2400
⇒2h/√3 = 2400
⇒ h = 1200√3 m

Q 17: For the statements p and q, consider the following compound statements:

(a) (~q∧(p→q)) → ~p

(b) ((p∨q))∧~p) → q

Then which of the following statements is correct?

(A) (A) is a tautology but not (B)

(B) (A) and (B) both are not tautologies

(C) (A) and (B) both are tautologies

(D) (B) is a tautology but not (A)

Answer: (C)

(A) is tautology.

(B) is tautology.
∴ (A) and (B) both are tautologies.

Q 18: Let a, b∈R. If the mirror image of the point P(a, 6, 9) with respect to the line (x – 3)/7 = (y – 2)/5 = (z – 1)/(-9) is (20, b, -a, -9), then |a + b| is equal to:

(A) 86

(B) 88

(C) 84

(D) 90

Answer: (B)
P(a, 6, 9), Q (20, b, -a-9)
Midpoint of PQ=((a+20)/2, (b+6)/2, -a/2) lie on the line.

=> (a + 20 – 6)/14 = (b + 6 – 4)/10 = (-a – 2)/(-18)
=> (a + 14)/14 = (a + 2)/18
=> 18a + 252 = 14a + 28
=> 4a = -224
a = -56
(b + 2)/10 = (a + 2)/18
=> (b + 2)/10 = (-54)/18
=>(b + 2)/10 = -3
=> b = -32
|a + b| = |-56 – 32|
= 88

Q 19: Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x∈R. If |f(x) f'(x) f'(x) f”(x)| = 0, for all x∈R, then the value of f(1) lies in the interval:

(A) (9, 12)

(B) (6, 9)

(C) (3, 6)

(D) (0, 3)

Answer: (B)
Given f(x) f”(x) – f'(x)² =0
Let h(x) = f(x)/f'(x)
Then h'(x) = 0
⇒ h(x) = k
⇒ f(x)/f'(x) = k
⇒ f(x) = k f’(x)
⇒ f(0) = k f'(0)
⇒ k = 1/2
Now, f(x) = ½ f'(x)
⇒∫ 2dx = ∫ f'(x)/f(x) dx
⇒ 2x = ln |f(x)| + C
As f(0) = 1 ⇒ C = 0
⇒ 2x = ln |f(x)|
⇒ f(x) = ±e^2x
As f(0) = 1 ⇒ f(x) = e^2x
∴ f(1) = e² ≈ 7.38

Q 20: A possible value of tan (¼ sin^-1 √63/8) is:

(A) 1/(2√2)

(B) 1/√7

(C) √7 – 1

(D) 2√2 – 1

Answer: (B)
tan (¼ sin^-1 √63/8)
Let sin^-1(√63/8) = θ
sin θ = √63/8

cos θ = 1/8
2 cos²(θ/2) – 1 = 1/8
⇒ cos² θ/2 = 9/16
cos θ/2 = 3/4
⇒(1- tan² θ/4 )/(1 + tan² θ/4) = 3/4
tan θ/4 = 1/√7

Section B

Q 1: If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x – 2)²+ (y – 3)² = 25 at the point (5, 7) is A, then 24A is equal to ___.

Answer: Q is wrong

Equation of normal at P is
(y – 7) = ((7 – 3)/(5 – 2))(x – 5)
⇒ 3y – 21 = 4x – 20
⇒ 4x – 3y + 1 = 0
⇒ M is (-1/4, 0)
Equation of tangent at P is
(y – 7) = (-¾) (x – 5)
⇒ 4y – 28 = -3x + 15
⇒ 3x + 4y = 43
⇒ N is (43/3, 0)
The question is wrong. The normal cuts at a point on the negative axis.

Q 2: If a + α = 1, b + β = 2 and af(x) + αf(1/x) = bx + β/2, x ≠ 0 then the value of the expression [f(x) + f(1/x) ]/(x + 1/x)

Answer: 2
af(x) + αf(1/x) = bx + β/x …(i)
Replace x by 1/x
af(1/x) + αf(x) = b/x + βx …(ii)
(i) + (ii)
(a + α)[f(x) + f(1/x) ]
= (x + 1/x)(b + β)
⇒ ( f(x) + f(1/x))/(x + 1/x)
= 2/1
= 2

Q 3: If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ________.

Answer: 11

⇒ 10(9 + k²) – (81 + k² + 18k) < 1000
⇒ 90 + 10k² – k² – 18k – 81 < 1000
⇒ 9k²– 18k + 9 < 1000
⇒ (k – 1)² < 1000/9
⇒ k – 1< (10√10)/3
⇒ k < (10√10)/3 + 1
Maximum possible integral value of k is 11.

Q 4: Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (-5, 0). If the locus of the point P is a circle of radius r, then 4r² is equal to ___.

Answer: 56.25
Let P be (h, k), A(5, 0) and B(-5, 0)
Given PA = 3PB
⇒ PA² = 9PB²
⇒ (h-5)² + k² = 9[(h + 5)² + k²]
⇒ 8h² + 8k² + 100h + 200 = 0
∴ Locus of P is x² + y² + (25/2)x + 25 = 0
Centre = (-25/4, 0)
∴ r² = (-25/4)2 – 25
= 625/16 – 25
= 225/16
∴4r² = 4×225/16
= 225/4
= 56.25

Q 5: The number of the real roots of the equation (x + 1)² + |x – 5| = 27/4 is ___.

Answer: 2
For x ≥ 5,
(x + 1)²+ (x – 5) = 27/4
⇒ x² + 3x – 4 = 27/4
⇒ x² + 3x – 43/4 = 0
⇒ 4x² + 12x – 43 = 0
x = (-12 ± √(144 + 688))/8
x = (-12 ± √832)/8
= (-12 ± 28.8)/8
= (-3 ± 7.2)/2
= (-3 + 7.2)/2, (-3 – 7.2)/2 (therefore, no solution)
For x < 5,
(x + 1)²– (x – 5) = 27/4
⇒ x² + x + 6 – 27/4=0
⇒ 4x² + 4x – 3=0
x = (-4 ± √(16 + 48))/8
x = (-4 ± 8)/8
⇒ x = -12/8, 4/8
∴ 2 real roots.

Q 6: The students S₁, S₂,… S₁₀ are to be divided into 3 groups A, B, and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___.

Answer: 31650

Number of ways
= ¹⁰C₁ [2⁹ – 2] + ¹⁰C₂ [2⁸– 2] + ¹⁰C₃ [2⁷ – 2]
= 27 [¹⁰C₁×4 + ¹⁰C₂ ×2 + ¹⁰C₃] – 20 – 90 – 240
= 128 [40 + 90 + 120 ] – 350
= (128 × 250) – 350
= 10 [3165]
= 31650

Q 7: For integers n and r, let

The maximum value of k for which the sum

exists, is equal to ____.

Answer: Bonus
(1+x)¹⁰ = ¹⁰C₀ + ¹⁰C₁x + ¹⁰C₂x² + …… + ¹⁰C₁₀x¹⁰
(1+x)¹⁵= ¹⁵C₀ + ¹⁵C₁x +¹⁵C₂x² + …… + ¹⁵C_k-1x^k-1 + ¹⁵C_k+1x^k+1 + …¹⁵C₁₅x¹⁵

Coefficient of x^k+1 in (1+x)²⁵
= ²⁵C_k+1
²⁵C_k + ²⁵C_k+1 = ²⁶C_k+1
For maximum value
as per given, k can be as large as possible.

Q 8: Let λ be an integer. If the shortest distance between the lines x – λ = 2y – 1 = -2z and x = y + 2λ = z – λ is √7/2√2, then the value of |λ| is __

Answer: 1
(x – λ)/1 = (y – 1/2)/(1/2) = z/(-1/2)
(x – λ)/2 = (y-1/2)/1 = z/(-1) …(1) Point on line = (λ, 1/2, 0)
x/1 = (y + 2λ)/1 = (z – λ)/1 …(2) Point on line = (0, -2λ, λ)

= |-5λ – 3/2|/√14
= √7/(2√2) (Given)
⇒ |10λ + 3| = 7
⇒ λ = -1 as λ is an integer.
⇒ |λ| = 1

Q 9: If i = √-1. If [(-1 + i√3)²¹/(1 – i)²⁴ + (1 + i√3)²¹/(1 + i)²⁴ ] = k, and n = [|k| ] be the greatest integral part of |k|. Then

is equal to ___

Answer: 310

= 2⁹ e^i(20π) +2⁹ e^iπ
= 2⁹ + 2⁹ (-1)
= 0 = k
∴ n = 0
∑(j=0)⁵ (j + 5)²– ∑(j=0)⁵ (j + 5)
= [5² + 6² + 7² + 8² + 9² + 10² ] – [5 + 6 + 7 + 8 + 9 + 10]
= [(1²+ 2² +… + 10²) – (1² + 2²+ 3²+ 4²) ] – [(1 + 2 + 3 + … + 10) – (1 + 2 + 3 + 4) ]
= (385 – 30) -[55 – 10]
= 355 – 45
= 310

Q 10: The sum of first four terms of a geometric progression (G.P.) is 65/12 and the sum of their respective reciprocals is 65/18. If the product of first three terms of the G.P. is 1, and the third term is α then 2α is____

Answer: 3
Let terms of GP are a, ar, ar², ar³
a + ar + ar² + ar³ = 65/12 …….(1)
1/a + 1/ar + 1/(ar²) + 1/ar³) = 65/18
⇒1/a (r³ + r² + r + 1)/r³) = 65/18 …….(2)
(1)/(2), we get
a²r³ = 18/12
= 3/2
Also, a³ r³ = 1
⇒ a(3/2) = 1
⇒a = 2/3
(4/9) r³ = 3/2
⇒ r³ = 33/23
⇒ r = 3/2
α = ar²
= 2/3.(3/2)² = 3/2
∴2α = 3

[/tab]

[/tabs]

[tabs title=”JEE MAINS EXAM Previous Year Paper” type=”centered”]

[tab title=”Physics”]

PHYSICS

SECTION A

Q 1. A current through a wire depends on time as i = α₀t + βt² where α₀ = 20 A/s and β = 8 As^-2. Find the charge crossed through a section of the wire in 15 s.

(A) 2100 C

(B) 260 C

(C) 2250 C

(D) 11250 C

Answer: (D) 

 

Q 2. Each side of a box made of metal sheet in cubic shape is ‘a’ at room temperature ‘T’, the coefficient of linear expansion of the metal sheet is ‘α’. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T+ΔT. Calculate the increase in the volume of the metal box:

(A) 4/3 πa³ α ΔT

(B) 4πa³ α ΔT

(C) 3a³ α ΔT

(D) 4a³ α ΔT

Answer: (C)

 

Q 3. Given below are two statements:

Statement-I: Two photons having equal linear momenta have equal wavelengths.

Statement-II: If the wavelength of a photon is decreased, then the momentum and energy of a photon will also decrease.

In the light of the above statements, choose the correct answer from the options given below.

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

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

(C) Both Statement-I and Statement-II are false

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

Answer: (D) 

 

Q 4. A cube of side ‘a’ has point charges +Q located at each of its vertices except at the origin where the charge is –Q. The electric field at the centre of cube is:

Answer: (C) 

 

Q 5. If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

Answer: (A)

 

Q 6. In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:

Answer: (A) 

 

Q 7. Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be

​

Answer: (A)

 

Q 8. In Young’s double-slit experiment, the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

(A) 4: 1

(B) 2: 1

(C) 3: 1

(D) 1: 4

Answer: (A) 

 

Q 9. Consider two satellites S₁ and S₂ with periods of revolution 1 hr and 8 hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellite S₂ is

(A) 8: 1

(B) 1: 8

(C) 2: 1

(D) 1: 4

Answer: (A)

 

Q 10. If an emitter current is changed by 4mA, the collector current changes by 3.5 mA. The value of β will be:

(A) 7

(B) 0.875

(C) 0.5

(D) 3.5

Answer: (A)

 

Q 11. n moles of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes:

A→ B: Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and pressure changes from P₁ to P₂.

B → C: Isobaric compression at pressure P₂ to initial volume V₁.

C → A: Isochoric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is –

(A) 0

(B) nRT(ln₂ + 1/2)

(C) nRTln₂

(D) nRT (ln₂ – 1/2)

Answer: (D)

 

Q 12. The work done by a gas molecule in an isolated system is given by, , where x is the displacement, k is the Boltzmann constant and T is the temperature α and β are constants. Then the dimensions of β will be:

 

(A) [M⁰LT⁰]

(B) [M²LT²]

(C) [MLT^–2]

(D) [ML²T^–2]

Answer: (C)

 

Q 13. Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be

(A) 2: 1

(B) 1: 4

(C) 4: 1

(D) 1: 2

Answer: (B)

 

Q 14. Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as:

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then:

(A) I₁ = I₂ = I₃< I₄

(B) I₁ + I₂ = I₃ + 5/2 I₄

(C) I₁ + I₃< I₂ + I₄

(D) I₁ = I₂ = I₃> I₄

Answer: (D)

 

Q 15. A cell E₁ of emf 6V and internal resistance 2Ω is connected with another cell E₂ of emf 4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is

(A) 3.6V

(B) 10.0V

(C) 5.6V

(D) 2.0V

Answer: (C) 

 

Q 16. The focal length f is related to the radius of curvature r of the spherical convex mirror by:

(A) f = r

(B) f = – ½ r

(C) f = +½ r

(D) f = – r

Answer: (C) 

 

Q 17. If Y, K and η are the values of Young’s modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

Answer: (A) 

 

Q 18. In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.

The transition A, B and C respectively represents:

(A) The series limit of Lyman series, third member of Balmer series and second member of Paschen series

(B) The first member of the Lyman series, third member of Balmer series and second member of Paschen series

(C) The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series

(D) The series limit of Lyman series, second member of Balmer series and second member of Paschen series

Answer: (A)

 

Q 19. Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is

Answer: (A) 

 

Q 20. Match List-I with List-II

List – I	List – II
(A)  Isothermal	(i) Pressure constant
(B)  Isochoric	(ii) Temperature constant
(C)  Adiabatic	(iii) Volume constant
(D)  Isobaric	(iv) Heat content is constant

Choose the correct answer from the options given below:

(A) (A)  – (ii), (B)  – (iv), (C)  – (iii), (D)  – (i)

(B) (A)  – (ii), (B)  – (iii), (C)  – (iv), (D)  – (i)

(C) (A)  – (i), (B)  – (iii), (C)  – (ii), (D)  – (iv)

(D) (A)  – (iii), (B)  – (ii), (C)  – (i), (D)  – (iv)

Answer: (B) 

SECTION-B

Q 1. In connection with the circuit drawn below, the value of current flowing through the 2kΩ resistor is _______ × 10–4 A.

Answer: 25
In zener diode there will be o change in current after 5V
Zener diode breakdown
⇒ i = 5 / 2 × 103
⇒ i = 2.5 × 10^–3 A
⇒ i = 25 × 10^–4 A

 

Q 2. An inclined plane is bent in such a way that the vertical cross-section is given by y = x² / 4 where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction μ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _____ cm.

Answer: 25

Given,
y = x² / 4
μ = 0.5
Condition for block will not slip downward
mg sin θ = μmg cos θ

⇒ tan θ = μ
And we know that
⇒ tanθ = dv / dx
⇒ dv / dx = μ ⇒ x/2 = 0.5 [y = x² / 4 dy / dx = x / 2]
⇒ x = 1,
put x = 1 in equation y = x²/4
⇒ y = (1)2 / 4 ⇒ y = ¼ ⇒ y = 0.25
y = 25 cm

 

Q 3. An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of the light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30° in clockwise direction, the intensity of emerging light will be________ Lumens.

 

Answer: 75
Given: I₀ = 100 lumens, θ = 30^o
Inet = I₀ cos2θ
Inet = 100 × (√3/2)2 = 100 × 3 / 4
Inet = 75 lumens

 

Q 4. The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ______ N.

[g = 10 ms-2]

Answer: 25

Given: μs = 0.2
m = 0.5 kg
g = 10 m/s2
We know that
fs = μsN and …. (1)
To keep the block adhere to the wall
Here, N = F … (2)
fs = mg …. (3)
From equation (1), (2), and (3), we get
⇒mg = μs F
⇒ F = mg / μs ⇒ F = 0.5 × 10 / 0.2
F = 25 N (for all values of F greater than or equal to 25 N this case is possible)

 

Q 5. A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift _______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ‘m’ on the smaller piston.

Answer: 25600

Atmospheric pressure P0 will be acting on both the limbs of the hydraulic lift.
Applying pascal’s law for the same liquid level
⇒ P₀ + mg / A₁ = P_o + (100)g / A₂
⇒ mg / A₁ = (100)g / A₂ ⇒ m / 100 = A₁/ A₂ …(1)
Diameter of piston on side of 100 kg is increased by 4 times so new area = 16A₂
Diameter of piston on side of (m) kg is decreasing
A₁ = A₁ / 16
(In order to increasing weight lifting capacity, diameter of smaller piston must be reduced)
Again, mg / (A₁/16) = M’g / 16A₂ ⇒ 256m / M’ = A₁/ A₂
From equation (1) = 256m / M’ = m / 100 ⇒ M’ = 25600 kg

 

Q 6. An audio signal υm = 20sin2π(1500t) amplitude modulates a carrier υc =80 sin 2π (100,000t). The value of percent modulation is ________.

Answer: 25
We know that, modulation index = A_m / A_c
From given equations, A_m = 20 and A_c = 80
Percentage modulation index = A_m / A_c × 100
⇒ 20 / 80 × 100 = 25%
The value of percentage modulation index is
= 25

 

Q 7. A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of the secondary coil are 24, then the number of turns of the primary are ______.

Answer: 440
Given,
Primary voltage, Vp = 220 V
Secondary voltage, vs = 12 V
No. of turns in secondary coil is Ns = 24
No. of turns in primary coil, Np = ?
We know that for a transformer
⇒ Np / Ns = Vp / Vs
⇒ Np = Vp × Ns / Vs = 220 × 24 / 12
⇒ Np = 440

 

Q 8. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30° with the original direction. The ratio of velocities of the balls after collision is x : y, where x is _________.

Answer: 1

Momentum is conserved just before and just after the collision in both x-y direction.
In y-direction,
pi = 0
pf = mv₁ sin30^o – mv₂ sin30^o
Pf = m × ½ v₁ – m × ½ v₂
pi = pf, so
= mv₁ / 2 – mv₂ / 2 = 0
⇒ mv₁ / 2 = mv₂ / 2 ⇒ v₁ = v₂
v₁ / v₂ = 1

 

Q 9. An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is _______ × 107 m/s.

Answer: 15
Given: f = 5 GHz
εr =2
μr = 2
Velocity of wave ⇒ v = c / n ….(1)
Where, n = √μrεr and c = speed of light = 3 × 108 m/s
n = √2 × 2 = 2
put the value of n in we get
⇒ v = 3 × 108 / 2 = 15 × 107 m/s
⇒ X × 107 = 15 × 107
X = 15

 

Q 10. A resonance circuit having inductance and resistance 2 × 10–4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of the quality factor of this resonator is________. [π = 3.14]

Answer: 2000
Given: R = 6.28 Ω
f = 10 MHz
L = 2 × 10-4 Henry
We know that quality factor Q is given by
⇒ Q = XL / R = ωL / R
also, ω = 2πf, so
⇒ Q = 2πfL / R
⇒ Q = 2π × 10 × 106 × 2 × 10-4 / 6.28 = 2000
Q = 2000

[/tab]

[tab title=”Chemistry”]

CHEMISTRY

SECTION-A 

Q 1. Which reagent (A) is used for the following given conversion?

(A) Cu / ∆ / high pressure

(B) Molybdenum Oxide

(C) Manganese Acetate

(D) Potassium Permanganate

Answer: (B)

 

Q 2. S-1: Colourless cupric metaborate is converted into cuprous metaborate in a luminous flame.

S-2: Cuprous metaborate is formed by reacting copper sulphate with boric anhydride heated in non luminous flame.

(A) S-1 is true and S-2 is false

(B) S-1 is false and S-2 is true

(C) Both are true

(D) Both are false

Answer: (D)

 

Q 3. Find A and B.

Answer: (B)

 

Q 4. EoM2+/ M has a positive value for which of the following elements of 3d transition series?

(A) Cu

(B) Zn

(C) Cr

(D) Co

Answer: (A)

 

Q 5. What is the reason for the formation of a meta product in the following reaction?

(A) Aniline is ortho/para directing

(B) Aniline is meta directing

(C) In acidic medium, aniline is converted into anilinium ion, which is ortho/para directing

(D) In acidic medium, aniline is converted into anilinium ion which is meta directing

Answer: (D) 

 

Q 6. Identify X, Y, Z in the given reaction sequence.

(A) X = Na[Al(OH)₄] ; Y = CO₂ ; Z = Al₂O₃.xH₂O

(B) X = Na[Al(OH)₄] ; Y = SO₂ ; Z = Al₂O₃.xH₂O

(C) X = Al(OH)₃ ; Y = CO₂ ; Z = Al₂O₃

(D) X = Al(OH)₃ ; Y = SO₂ ; Z = Al₂O₃

Answer: (A) 

 

Q 7. The missing reagent P is:

Answer: (A) 

Q 8. Arrange Mg, Al, Si, P and S in the correct order of their ionisation potentials.

 

Answer: P > S > Si > Mg > Al

 

Q 9. Which force is responsible for the stacking of the α-helix structure of protein? Hydrogen (¹H, ²H, ³H) is ________.

(A) H-bond

(B) Ionic bond

(C) Covalent bond

(D) Van der Waals forces

Answer: (A) 

 

Q 10. The composition of gun metal is:

(A) Cu, Zn, Sn

(B) Al, Mg, Mn, Cu

(C) Cu, Ni, Fe

(D) Cu, Sn, Fe

Answer: (A)

 

Q 11. The gas evolved due to anaerobic degradation of vegetation causes:

(A) Global warming and cancer

(A) Global warming and cancer

(B) Acid rain

(C) Ozone hole

(D) Metal corrosion

Answer: (A) 

 

Q 12. The slope of the straight line given in the following diagram for adsorption is:

(A) 1/n (0 to 1)

(B) 1/n (0.1 to 0.5)

(C) log n

(D) log (1/n)

Answer: (A)

 

Q 13. Match the following:

(i) Caprolactam	(a)  Neoprene
(ii) Acrylonitrile	(b)  Buna N
(iii) 2-chlorobuta-1,3-diene	(c)  Nylon – 6
(iv) 2-Methylbuta-1,3-diene	(d)  Natural rubber

(A) (i) →(b) , (ii) → (c) , (iii) → (a) , (iv) → (d) 

(B) (i) → (a) , (ii) → (c) , (iii) → (b) , (iv) → (d) 

(C) (i) → (c) , (ii) → (b) , (iii) → (a) , (iv) → (d) 

(D) (i) → (c) , (ii) → (a) , (iii) → (b) , (iv) → (d) 

Answer: (C)

 

Q 14. In the given reactions,

1) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂

2) H₂O₂ + HOCl → Cl^– + H₃O+ + O₂

(A) H₂O₂ acts as an oxidising agent in both the reactions

(B) H₂O₂ acts as a reducing agent in both the reactions

(C) H₂O₂ acts as an oxidising agent in reaction (1) and as a reducing agent in reaction (2)

(D) H₂O₂ acts as a reducing agent in reaction (1) and as an oxidizing agent in reaction (2)

Answer: (B) 

 

Q 15. What is the major product of the following reaction

Answer: (A) 

 

Q 16. Which of the following ores are concentrated by cyanide of group 1st element?

(A) Sphalerite

(B) Malachite

(C) Calamine

(D) Siderite

Answer: (A)

 

Q 17. What is the major product of the following reaction?

Answer: (C) 

 

Q 18. Which of the following pairs are isostructural

a) TiCl₄, SiCl₄

b) SO^2-₃, CrO^2-₃

c) NH₃, NO^–₃

d) ClF₃, BCl₃

(A) B, C

(B) A, C

(C) A, B

(D) A, D

Answer: (C)

 

Q 19. Identify the major product:

Answer: (B) 

Q 20. The products A and B are:

 

Answer: (A)

Section B

Q 1. Cu²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ K₁ = 10⁴

[Cu(NH₃)]²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ K₂ = 1.58 × 10³
[Cu(NH₃)₂]²⁺ + NH₃ ⇌ [Cu(NH₃)₃]²⁺ K₃ = 5 × 10²
[Cu(NH₃)₃]²⁺ + NH₃ ⇌ [Cu(NH₃)₄]²⁺ K₄ = 10²
If the dissociation constant of [Cu(NH₃)₄]²⁺ is X × 10¹². 

 

Determine X.
Answer: 1.26
Overall reaction constant (β):
β = K₁× K₂ × K₃ × K₄
= 104 (1.58 ×103) × 5 × 102 × 102 = 7.9 × 10¹¹
Cu₂+ + 4NH₃ ⇌ [Cu(NH₃)₄]²⁺ ; β = 7.9 × 10¹¹
So, the dissociation constant (K_Disso.) will be:

1/β = 1/ 7.9 × 10¹¹

​K_Disso = 1 / β = 1.26 × 10^-12
Hence, the value of X = 1.26

 

Q 2. What is the coordination number in Body Centered Cubic (BCC) arrangement of identical particles?

Answer: 8
The easiest way is to look at the atom at the body center. It lies at the center of the body diagonal and touches the eight corner atoms. So, coordination number = 8.

 

Q 3. Cl₂(g)⇌ 2Cl(g)

For the given reaction at equilibrium, moles of Cl₂(g) is equal to the moles of Cl(g) and the equilibrium pressure is 1atm. If Kp of this reaction is x ×10^–1, find x.

Answer: 5
According to the question: Cl₂ ⇌ 2Cl [P_total = 1 atm].
Moles of Cl₂ = Moles of Cl (at equilibrium).
Given: Kp = x ×10^-1
nCl₂ = nCl.
Therefore, P Cl₂ = P Cl
Hence, P Cl₂ = P Cl = 0.5 atm
kp= (P Cl)₂/ P Cl₂ = 0.5 = 5 x 10^-1
So, x = 5

 

Q 4. Among the following compounds, how many are amphoteric in nature?

Be(OH)₂, BeO, Ba(OH)₂, Sr(OH)₂

Answer: 2
The oxide and hydroxide of Be are amphoteri(C) Hence BeO and Be(OH)2 is amphoteri(C) Ba(OH)₂ and Sr(OH)₂ are basic.

 

Q 5. S₈ + bOH^– → cS^2- + sS₂O^2-₃ + H₂O. Find the value of c.

Answer: 4
Let us look at the half reactions i.e. oxidation and reduction separately.
Oxidation:
S₈ → S₂O^2-₃
S₈ + 24OH^– → 4S₂O^2-₃ + 12H₂O +16e- ….(1)
Reduction:
S₈ → S^2-
S₈ + 16e- → 8S^2- ….(2)
Adding both the reactions (1) and (2),
2S₈ + 24OH^– → 4S₂O^2-₃ + 8S^2- + 12H₂O
Dividing the whole equation by 2,
S₈ + 12OH^– → 2S₂O^2-₃ + 4S^2- + 6H₂O
So, c = 4

 

Q 6. 4.5 g of a solute having molar mass of 90 g/mol is dissolved in water to make a 250 mL solution. Calculate the molarity of the solution.

Answer: 0.2
W_B (given weight of solute) = 4.5 g
M_B (Molar mass of solute) = 90 g/mol
V_S (Volume of solution) = 250 mL
= 250 / 1000 L = 1/4 L
Molarity (M) = nB / V_S(L) = W_B / M_B.V_S(L) = 4.5 / (90 ×1/4) = 0.2 molar

 

Q 7. Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is 3.3× 10-4 sec-1.

Answer: 1518

= 30 x 103 x 0.506 = 1518 sec

 

Q 8. 9.45g of CH₂ClCOOH is dissolved in 500 mL of H₂O solution and the depression in freezing point of the solution is 0.5°C. Find the percentage dissociation.

(Kf)H₂O = 1.86K kg mole^-1.

Answer: 34.4%
CH₂Cl COOH CH₂ClCOO^– + H⁺
Initial 100
Dissociated α α α
Left (1-α) α α
i = final moles / initial moles = 1 – α + α + α / 1 = 1 + α
ΔTf = i × K_f × m
0.5 = (1 + α) × 1.86 × m
Molality (m) = nB / WA (kg) = WBMB x WA (kg) = 9.45 × 1000 / 94.5 × 500 = 0.2
Here, A is for solute and B is for solvent
W_A=W_H2O = 500g (Density of H₂O = 1g/mL)
0.5 = (1+ α) ×1.86 × 0.2
α = 0.344
Percentage of dissociation = 34.4%

 

Q 9. For a chemical reaction, Keq is 100 at 300K, the value of ΔGo is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)

Answer: 1382
Given: K_eq= 100 at 300K and ΔG^o = –xR Joule
ΔGo = -RTln(K_eq) = -2.303RTlog(K_eq) = -1381.8R
Therefore, x = 1381.8 or 1382

 

Q 10. The mass of Li³⁺ is 8.33 times the mass of a proton. If Li3+ and proton are accelerated through the same potential difference, then the ratio of de Broglie’s wavelength of Li³⁺ to proton is x ×10^–1. Find x

Answer: 2

m (Li³⁺) = 8.33 × m_p+ (given)
Debroglie’s wavelength (λ) = h / p
KE = ½ mv²
Multiplying by m on both sides
m. KE = ½ mv²
2m. KE = (mv)² = p²
P = √2m. KE
Also KE= q × V
So, P = 2mq.V

Comparing with x × 10 ^-1 = 2 × 10 ^-1
x = 2

[/tab]

[tab title=”Maths”]

MATHS 

SECTION A

Q 1: The statement among the following that is a tautology is:

(A) A∧(A∨B)

(B) B→[A∧(A→B)]

(C) A∨(A∧B)

(D) [A∧(A→B)]→B

Answer: (D)
A∧ (~ A∨B)→B
= [(A∧~A)∨(A∧B)]→ B
= (A∧ B)→ B
= ~ (A∧B)∨B
= t

 

Q 2: Let f :R→R be defined as f(x) = 2x-1 and g:R – {1} →R be defined as g(x) = (x-½)/(x-1).

Then the composition function f(g(x)) is:

(A) Both one-one and onto

(B) onto but not one-one

(C) Neither one-one nor onto

(D) one-one but not onto

Answer: (D)
f(g(x)) = 2g(x) – 1
= 2(x – 1/2)/(x – 1) – 1
= x/(x – 1)
f(g(x)) = 1 + 1/(x – 1)

∴ one-one, into

 

Q 3: If f:R→ R is a function defined by f(x) = [x – 1] cos ((2x – 1)/2)𝜋 , where [.] denotes the greatest integer function, then f is:

(A) discontinuous only at x = 1

(B) discontinuous at all integral values of x except at x = 1

(C) continuous only at x = 1

(D) continuous for every real x

 

Answer: (D)
Doubtful points are x = n, n∈I

f(n) = 0
Hence continuous.

 

Q 4: If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

(A) –2t³

(B) –t³

(C) 0

(D) 2t³

Answer: (A)
Equation of tangent at P(t, t³)
(y – t³) = 3t²(x – t) ⋯(1)
Now solve the above equation with
y = x³ ⋯(2)
By (1) & (2)
x³ – t³ = 3t² (x – t)
x² + xt + t² = 3t²
x² + xt – 2t² = 0
⇒(x – t)(x + 2t) = 0
⇒x = – 2t
⇒Q(-2t, -8t3)
Ordinate of required point = (2t³ + (-8t)³)/3
= -2t³

 

Q 5: The value of – ¹⁵C₁ + 2. ¹⁵C₂ – 3.¹⁵C₃+ ….-15.¹⁵C₁ + ¹⁴C₁+ ¹⁴C₃ + ¹⁴C₅ + …¹⁴C₁₁ is

(A) 2¹⁴

(B) 2¹³ – 13

(C) 2¹⁶ – 1

(D) 2¹³ – 14

Answer: (D) 

= (¹⁴C₁+ ¹⁴C₃ + ¹⁴C₅ + …¹⁴C₁₁ + ¹⁴C₁₃) – ¹⁴C₁₃
= 2¹³ – 14
∴ S₁ + S₂ = 2¹³ – 14

 

Q 6: An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:

(A) 3/16

(B) 1/2

(C) 5/16

(D) 1/32

Answer: (B)
P(odd no. twice) = P(even no. thrice)
ⁿC₂ (1/2)ⁿ = ⁿC₃ (1/2)ⁿ
⇒ n = 5
Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)
= ⁵C₂ (1/2)⁵ + ⁵C₂ (1/2)⁵ + ⁵C₂ (1/2)⁵
= 16/2⁵
= 1/2

 

Q 7: Let p and q be two positive number such that p + q = 2 and p4 + q4 = 272. Then p and q are roots of the equation:

(A) x² – 2x + 2 = 0

(B) x² – 2x + 8 = 0

(C) x² – 2x + 136 = 0

(D) x² – 2x + 16 = 0

Answer: (D)
(p² + q²)² – 2p²q² = 272
((p + q)² – 2pq)² – 2p²q² = 272
⇒16 – 16pq + 2p²q² = 272
(pq)² – 8pq –128 = 0
⇒pq = (8±24)/2 = 16, – 8
⇒pq = 16
Now
x² – (p + q)x + pq = 0
x² – 2x + 16 = 0

 

Q 8: The area (in sq.units) of the part of the circle x² + y² = 36, which is outside the parabola y² = 9x, is:

(A) 24π + 3√3

(B) 12π + 3√3

(C) 12π – 3√3

(D) 24π – 3√3

 

Answer: (D)
The curves intersect at point (3, ± 3√3)

Required area

 

Q 9: If ∫(cos x -sin x)/√(8-sin 2x) dx = a sin-1(sin x + cos x)/b + c where c is a constant of integration, then the ordered pair (a, b) is equal to:

(A) (1, –3)

(B) (1, 3)

(C) (–1, 3)

(D) (3, 1)

 

Answer: (B)
Put sin x + cos x = t ⇒1 + sin 2x = t²
(cos x -sin x ) dx = dt
⇒I = ∫dt/√(8-(t²-1))
= ∫dt/(9-t²)
= sin^-1 (t/3) + c = sin^-1(sin x + cos x)/3 + c
⇒ a = 1 and b = 3

 

Q 10: The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is:

(A) x = a

(B) x = 0

(C) x = -a/2

(D) x = a/2

Answer: (B)

h = (at²+a)/2, k = (2at+0)/2
⇒ t² =(2h-a)/a and t=k/a
⇒ k²/a² =(2h-a)/a
⇒ Locus of (h, k) is y² = a (2x – a)
⇒ y² = 2a(x- a/2)
Its directrix is x – a/2 = -a/2
⇒ x = 0

 

Q 11: A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 1/4. Three stones A, B and C are placed at the points (1,1),(2,2), and (4,4) respectively. Then which of these stones is/are on the path of the man?

(A) B only

(B) A only

(C) the three

(D) C only

 

Answer: (A)
x/a + y/b = 1
h/a + k/b = 1 ⋯⋯(1)
and (1/a + 1/b)/2 = 1/4
∴1/a + 1/b = 1/2 ⋯⋯(2)
(From (1) and (2))
Line passes through fixed point B(2, 2)

 

Q 12: The function f(x) = (4×3- 3×2)/6 – 2sin x + (2x – 1)cos x:

(A) increases in [1/2, ∞)

(B) decreases (-∞, 1/2]

(C) increases in (-∞, 1/2]

(D) decreases [1/2, ∞)

Answer: (A)
f’(x) = (2x – 1) (x – sin x )
⇒ f’x ≥ 0 in x∈(-∞, 0] ⋃ [1/2, ∞)
and f’x ≤ 0 in x∈(0, ½)

 

Q 13: The distance of the point (1, 1, 9) from the point of intersection of the line (x – 3)/1 = (y – 4)/2 = (z – 5)/2 and the plane x + y + z = 17 is:

(A) 38

(B) 19√2

(C) 2√19

(D) √38

Answer: (D)
(x – 3)/1 = (y – 4)/2 = (z – 5)/2 = λ
x = λ + 3, y = 2λ+ 4, z = 2λ+5
Which lies on given plane hence
⇒ λ+3+2λ +4+2λ+5 = 17
⇒ λ = 5/5 = 1
Hence, point of intersection is Q (4, 6, 7)
∴ Required distance =PQ
= √(9+25+4)
= √38

 

Q 14:

(A) 2/3

(B) 0

(C) 1/15

(D) 3/2

 

Answer: (A)

 

Q 15: Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

(A) 25

(B) 20√3

(C) 30

(D) 25√3

 

Answer: (D)

tan θ = h/75 = 75/3h
h² = 75²/3
h = 25√3m

 

Q 16: A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

(A) 560

(B) 1050

(C) 1625

(D) 575

Answer: (C)
(2I,4F)+ (3I,6F) + (4I,8F)
= ⁶C₂ ⁸C₄ + ⁶C₃ ⁸C₆ + ⁶C₄ ⁸C₈​

= 15 × 70 + 20 × 28 + 15 × 1
= 1050 + 560 + 15 = 1625

 

Q 17: The equation of the plane passing through the point (1,2,–3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is:

(A) 3x – 10y – 2z + 11 = 0

(B) 6x – 5y – 2z – 2 = 0

(C) 11x + y + 17z + 38 = 0

(D) 6x – 5y + 2z + 10 = 0

Answer: (C)

 

Q 18: The population P = P(t) at time ‘t’ of a certain species follows the differential equation dP/dt = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is:

(A) ½ log_e 18

(B) 2log_e 18

(C) log_e 9

(D) log_e 18

Answer: (B)
dp/dt = (p-900)/2

ln |900| – ln |50| = t/2
t/2 = ln |18|
⇒ t = 2ln 18

 

Q 19: If e^{(cos²x + cos⁴x + cos⁶x + …..∞ )log_e2 satisfies the equation t² – 9t + 8 = 0, then the value of 2sin x/(sin x + √3cos x) 0< x<π/2 is:

(A) 3/2

(B) 2√3

(C) 1/2

(D) √3

 

Answer: (C)

0 < x < π/2 ⇒ cot⁡x = √3
⇒(2 sin⁡x)/(sin⁡x+√3 cos⁡x )
= 2/(1+√3 cot ⁡x)
= 1/2

 

Q 20: The system of linear equations

3x – 2y – kz = 10

2x – 4y – 2z = 6

x + 2y – z = 5m

is inconsistent if :

(A) k = 3, m = 4/5

(B) k ≠ 3,m∈R

(C) k ≠ 3, m ≠ 4/5

(D) k = 3, m ≠ 4/5

 

Answer: (D)

⇒ 3(4 + 4) + 2(–2 + 2) -k(4 + 4) = 0
⇒ k = 3

⇒ 10(4 + 4) +2(-6 + 10m) -3(12 + 20m) ≠ 0
⇒ m ≠ 4/5

⇒ 3(-6 + 10m) – 10(- 2 + 2) – 3(10m – 6) ≠ 0
⇒ 0

⇒3(-20m – 12) +2(10m – 6) +10(4 + 4) ≠ 0
⇒ m ≠ 4/5

Section B

Q 1:

Solution:

Answer: 1

= tan 𝜋/4
= 1

Q 2: If and [x] denotes the greatest integer ≤ x, then 

is equal to

Answer: 3

 

Q 3: If one of the diameters of the circle x²+y² – 2x – 6y + 6 = 0 is a chord of another circle ‘C’ whose center is at (2,1), then its radius is ________

Solution:

Answer: 3

distance between (1,3) and (2,1) is √5
∴ (√5)2+(2)2= r2
⇒r = 3

 

Q 4: Let three vectors a,  b and c be such that is coplanar with a and b, a.c=7 and is perpendicular to c, where and , then the value of is _______

 

Answer: 75

 

Q 5: The minimum value of α for which the equation 4/sin⁡x +1/(1-sin⁡x )=α has at least one solution in (0,π/2) is _______

Solution:

Answer: 9
f(x⁡) = 4/sin⁡x + 1/(1 – sin⁡x )
Let sin⁡x = t ∵x∈(0, π/2) ⇒ 0 < t < 1
f(t⁡)= 4/t + 1/(1-t)
f'(t⁡) = (-4)/t2 + 1/(1 – t⁡)2 = 0
⇒(t2-4(1-t⁡)2/t2(1-t⁡)2 = 0
⇒t = 2/3
f_min at t = 2/3
α_min = f(2/3) =4/(2/3) + 1/(1 – 2/3)
= 6 + 3
= 9

 

Q 6: Let A = {n∈N∶ n is a 3-digit number}, B = {9k + 2∶ k∈N} and C = {9k + l∶ k∈N} for some l(0 < l < 9). If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to ___________

Answer: 5
3 digit number of the form 9K+2 are {101,109,⋯,992}
⇒ Sum equal to (100/2)(1093) = S₁= 54650
Now 274 × 400 =S₁+S₂
⇒274 × 400 = (100/2) [101 + 992] + S₂
⇒274 × 400 = 50 × 1093 + S₂
⇒S₂ = 109600 – 54650
∴S₂ = 54950
S₂ = 54950 = (100/2) [(99+l) + (990+l)]
⇒ 2l + 1089 = 1099
⇒ l = 5

 

Q 7: Let

where α∈R. Suppose Q = [q_ij ] is a matrix satisfying PQ = kI₃ for some non-zero k∈R. If q₂₃= – k/8 and |Q⁡| = k²/2, then α² + k² is equal to ___

 

Answer: 17

 

Q 8: Let M be any 3×3 matrix with entries from the set {0,1,2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is __________

Answer: 540

⇒ a² + b² + c² + d² + e² + f² + g² + h² + i² = 7
Case I∶ Seven (1’s) and two (0’s)
⁹C₂ = 36
Case II∶ One (2) and three (1’s) and five (0’s)
9!/5!3! = 504
∴Total = 540

 

Q 9: If the least and the largest real values of α, for which the equation z + α|z⁡-1| + 2 i⁡ = 0 (z∈C and i =√(-1)) has a solution, are p and q respectively; then 4(p2 + q2) is equal to _________

 

Answer: 10
x + iy + α√((x–1)² +y²) + 2i=0
⇒y + 2 = 0 and x + α√((x-1)²+y²)=0
y = –2 & x² = α²(x² – 2x + 1 + 4)
α2 = x²/(x²– 2x + 5)
⇒ x² (α² – 1) – 2xα² + 5α² = 0
∵x∈R ⇒ D≥0
⇒ 4α⁴ – 4(α² – 1)5α² ≥ 0
⇒ α² [4α² – 20α² + 20] ≥ 0
⇒ α² [-16α² + 20] ≥ 0
⇒ α² [α² – 5/4] ≤ 0
⇒ α² ∈ [0, 5/4]
⇒ α ∈ [-√5/2, √5/2]
then 4[p² + q²] = 4[5/4 + 5/4]
= 10

 

Q 10: Let Bi (i = 1,2,3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α – 2β)p ⁡= αβ and (β – 3γ)p⁡= 2βγ (All the probabilities are assumed to lie in the interval (0, 1)). Then (P(B1⁡))/(P(B3⁡) )is equal to _____

 

Answer: 6
Let x,y,z be probability of B1, B2, B3 respectively.
⇒ x(1 – y) (1 – z) = α
⇒ y(1 – x)(1 – z) = β
⇒ z(1 – x)(1 – y ) = γ
⇒ (1 – x)(1 – y)(1 – z) = p
Now (α – 2β)p = αβ
⇒ (x(1–y)(1–z)-2y(1-x)(1–z)) (1–x)(1–y)(1–z) = xy(1–x)(1–y) (1–z)2
⇒ x+ xy – 2y = xy
∴x = 2y⋯(1)
Similarly, (β–3γ) p = 2βγ
⇒ y = 3z ⋯(2)
From (1) & (2)
⇒x = 6z
Hence x/z = P(B1)/P(B3) = 6

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