FINAL JEE- MAIN EXAMINATION - JULY, 2022  (Held On Friday 29th July, 2022) TIME : 9 : 00 AM to 12 : 00 NOON

PHYSICS

SECTION-A

1. Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S),

if density of the liquid is p and radius of the drop is r, then T = k√(pr³/s³) is dimensionally correct, where K is dimensionless.
Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(C) (A) is true but (R) is false
(D) (A) is false but (R) is true
Official Ans. by NTA (D)
Sol. T = k√(pr³/s³)
Dimensions of RHS = [M L?³] [L³] / [M T?²] = M? L? T²
Dimensions of L.H.S ≠ Dimensions of R.H.S
∴ option (D)

TEST PAPER WITH SOLUTION

2. A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h/3 while going up and coming down respectively.
(A) (√2 - 1)/(√2 + 1)
(B) (√3 - √2)/(√3 + √2)
(C) (√3 - 1)/(√3 + 1)
(D) 1/3
Official Ans. by NTA (B)
Sol. Max. Height = h = u²/2g
⇒ u = √(2gh)
S = ut + 1/2 at²
h/3 = √(2gh)t + 1/2(-g)t²
gt²/2 - √(2gh)t + h/3 = 0 (Roots are t? & t?)
t?/t? = (√(2gh) + √(2gh - 4×g/2×h/3)) / (√(2gh) - √(2gh - 4×g/2×h/3)) = (√3 + √2)/(√3 - √2)

3. If t = √x + 4, then (dx/dt)_{t=4} is:
(A) 4
(B) Zero
(C) 8
(D) 16
Official Ans. by NTA (B)
Sol. t = √x + 4
⇒ x = (t - 4)² = t² - 8t + 16
⇒ dx/dt = 2t - 8
⇒ dx/dt|_{t=4} = 2×4 - 8 = 0

4. A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass m moves against the wall with a speed v.

Which of the following curve represents the correct relation between the normal reaction on the block by the wall (N)

and speed of the block (v)?
Official Ans. by NTA (A)
Sol. N = mv²/r Curve is parabola Y = kx²

5. A ball is projected with kinetic energy E, at an angle of 60° to the horizontal. The kinetic energy of this ball at the highest point of its flight will become:
(A) Zero
(B) E/2
(C) E/4
(D) E
Official Ans. by NTA (C)
Sol. E = 1/2 mu²
At Highest point, Velocity V = u cos60° = u/2
∴ K.E. at topmost point = 1/2 m(u/2)² = E/4

6. Two bodies of mass 1 kg and 3 kg have position vectors i + 2j + k and -3i - 2j + k respectively.

The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:
(A) i - 2j + k
(B) -3i - 2j + k
(C) -2i + 2k
(D) -2i - j + k
Official Ans. by NTA (A)
Sol. (m?r? + m?r?)/(m? + m?) = (1(i + 2j + k) + 3(-3i - 2j + k))/(1 + 3) = -2i - j + k
|-2i - j + k| = √((-2)² + (-1)² + (1)²) = √6

7. Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Clothes containing oil or grease stains cannot be cleaned by water wash.
Reason (R) : Because the angle of contact between the oil/ grease and water is obtuse.
In the light of the above statements, choose the correct answer from the option given below.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(C) (A) is true but (R) is false
(D) (A) is true but (R) is true
Official Ans. by NTA (A)
Sol. θc > 90°
For water oil interface

8. If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modules of the material of the wire will :
(A) Remains same
(B) Become 8 times its initial value
(C) Become 1/4th of its initial value
(D) Become 4 times its initial value
Official Ans. by NTA (A)
Sol. Y depends on material of wire

9. The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α is given by :
(A) 2π√(L/(g cos α))
(B) 2π√(L/(g sin α))
(C) 2π√(L/g)
(D) 2π√(L/(g tan α))
Official Ans. by NTA (A)
Sol. g_eff = g cos α

10. A spherically symmetric charge distribution is considered with charge density varying as
Where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :
(A) ρ?r/4ε? (3/4 - r/R)
(B) ρ?r/3ε? (3/4 - r/R)
(C) ρ?r/4ε? (1 - r/R)
(D) ρ?r/5ε? (1 - r/R)
Official Ans. by NTA (C)
Sol. By Gauss law
? E·dS = Q_in/ε?
E·4πr² = ∫?? ρ?(3/4 - r/R)4πr² dr / ε?
E·4πr² = ρ?·4π(3/4 · r³/3 - r?/4R) / ε?
Er² = ρ?r³/4ε? {1 - r/R}
E = ρ?r/4ε? {1 - r/R}

11. Given below are two statements.
Statement I : Electric potential is constant within and at the surface of each conductor.
Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
In the light of the above statements, choose the most appropriate answer from the options give below.
(A) Both statement I and statement II are correct
(B) Both statement I and statement II are incorrect
(C) Statement I is correct but statement II is incorrect
(D) Statement I is incorrect but and statement II is correct
Official Ans. by NTA (A)
Sol. (Properties of conductor)
Statement - I, true as body of conductor acts as equipotential surface.
Statement - 2 True, as conductor is equipotential. Tangential component of electric field should be zero. Therefore electric field should be perpendicular to surface.

12. Two metallic wires of identical dimensions are connected is series. If σ? and σ? are the conductivities of the these wires respectively, the effective conductivity of the combination is :
(A) σ?σ?/(σ? + σ?)
(B) 2σ?σ?/(σ? + σ?)
(C) (σ? + σ?)/(2σ?σ?)
(D) (σ? + σ?)/(σ?σ?)
Official Ans. by NTA (B)
Sol. Let length of wire be ?
Area of wire as 'A'
For equivalent wire length = 2? & area will be A
Thermal resistance
R_eq = R? + R?
2?/(σ_eq A) = ?/(σ? A) + ?/(σ? A)
2?/(σ_eq A) = ?/σ? + ?/σ? ⇒ σ_eq = 2σ?σ?/(σ? + σ?)

13. An alternating emf E = 440 sin 100πt is applied to a circuit containing an inductance of √2/π H. If an a.c. ammeter is connected in the circuit, its reading will be :
(A) 4.4A
(B) 1.55A
(C) 2.2A
(D) 3.11A
Official Ans. by NTA (C)
Sol. E = 440 sin 100πt, L = √2/π H
X_L = ωL = 100π(√2/π) = 100√2 Ω
Peak current I? = E?/X_L = 440/(100√2) = 2.2√2 A
AC ammeter reads RMS value therefore reading will be I_rms
I_rms = I?/√2 = 2.2 A

14. A coil of inductance 1 H and resistance 100Ω is connected to a battery of 6 V. Determine approximately :
(a) The time elapsed before the current acquires half of its steady - state value
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln2 = 0.693, e^(-3/2) = 0.25)
(A) t = 10 ms; U = 2 mJ
(B) t = 10 ms; U = 1 mJ
(C) t = 7 ms; U = 1 mJ
(D) t = 7 ms; U = 2 mJ
Official Ans. by NTA (C)
Sol. Given circuit is R - L growth circuit
i = E/R (1 - e^(-t/τ))
i = E/(2R) = E/R (1 - e^(-t/τ))
Solving t = τ ln2
t = (1/R) ln2 = (1/100)(0.693) = 0.00693 s = 7 ms
i(15ms) = E/R (1 - e^(-15/10))
i = 6/100 (1 - 1/√4) = 3/4 × 6/100
U = 1/2 Li²,
by solving we get U = 1 mJ

15. Match List - I with List - II
List - I
(a) UV rays
(b) X-rays
(c) Microwave
(d) Infrared wave
List - II
(i) Diagnostic tool in medicine
(ii) Water purification
(iii) Communication, Radar
(iv) Improving visibility in foggy days
Choose the correct answer from the options given below :
A) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B) (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
C) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Official Ans. by NTA (B)
Sol. (a) uv rays - used for water purification
(b) x-rays used for diagnosing fracture
(c) Microwaves are used for mobile and radar communication
(d) Infrared waves show less scattering therefore used in foggy days
(a-ii), (b-i), (c-iii), (d-iv)

16. The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength :
(A) hc/(Eλ - hc)
(B) hcλ/(Eλ + hc)
(C) hλ/(Eλ + hc)
(D) hλ/(Eλ - hc)
Official Ans. by NTA (B)
Sol. E = hc/λ - φ ...(i)
2E = hc/λ' - φ ...(ii)
E = hc(1/λ - 1/λ')
⇒ λ' = hcλ/(Eλ + hc)

17. Find the ratio of energies of photons produced due to transition of an election of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
(A) 3:4
(B) 4:3
(C) 1:4
(D) 4:1
Official Ans. by NTA (A)
Sol. E_n = -13.6/n² eV
⇒ (E? - E?)/(E_∞ - E?) = 13.6(1 - 1/4)/13.6 = 3/4

18. Find the modulation index of an AM wave having 8 V variation where maximum amplitude of the AM wave is 9 V.
(A) 0.8
(B) 0.5
(C) 0.2
(D) 0.1
Official Ans. by NTA (A)
Sol. Modulation index: m = A_m/A_c
Given 2A_m = 8
A_m + A_c = 9 ⇒ A_c = 5
∴ m = 4/5 = 0.8

19. A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?
(A) 0.001 cm
(B) 0.002 mm
(C) 0.002 cm
(D) 0.005 cm
Official Ans. by NTA (C)
Sol. 1 MSD = 1/20 cm
1 VSD = 24/25 MSD = 24/25 × 1/20 cm
∴ Least count = 1/20(1 - 24/25) cm
= 1/20 × 1/25 = 1/500 cm
= 0.002 cm

20. In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:
(a) Screw moves 0.5 mm on main scale in one complete rotation
(b) Total divisions on circular scale = 50
(c) Main scale reading is 2.5 mm
(d) 45th division of circular scale is in the pitch line
(e) Instrument has 0.03 mm negative error
Then the diameter of wire is:
(A) 2.92 mm
(B) 2.54 mm
(C) 2.98 mm
(D) 3.45 mm
Official Ans. by NTA (C)
Sol. MSR = 2.5 mm
CSR = 45 × 0.5/50 mm = 0.45 mm
Diameter reading = MSR + CSR - zero error
= 2.5 + 0.45 - (-0.03) = 2.98 mm

SECTION-B

1. An object is projected in the air with initial velocity u at an angle θ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be degree.
Official Ans. by NTA (15)
Sol. R_max = u² sin 2(45°)/g = u²/g
R/2 = u²/2g = u² sin 2θ/g
sin 2θ = 1/2
2θ = 30°, 150°
θ = 15°, 75°
Ans. 15, 75

2. If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth αh (h << R_e) from the earth surface. The value of α will be
(use R_e = 6400 km)
Official Ans. by NTA (2)
Sol. g(1 - 2h/R) = g(1 - d/R)
2h/R = d/R
αh = d
α = 2

3. The pressure P? and density d? of diatomic gas (γ = 7/5) changes suddenly to P?(>P?) and d? respectively during an adiabatic process. The temperature of the gas increases and becomes ______ times of its initial temperature. (given d?/d? = 32)
Official Ans. by NTA (4)
Sol. PV^γ = const
d = m/v
p(m/d)^γ = const
p/d^γ = const
d?/d? = 32
p?/p? = (d?/d?)^γ = (1/32)^γ = 1/128
T?/T? = p?V?/(p?V?) = 1/128 × 32 = 1/4
T? = 4T?

4. One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α²/4 R J/mol K; then the value of α will be (Assume that the given diatomic gas has no vibrational mode.)
Official Ans. by NTA (3)
Sol. C_v mix = (n?C_v? + n?C_v?)/(n? + n?)
= (1 × 3R/2 + 3 × 5R/2)/4 = 9R/4 = α²/4 R
α = 3

5. The current I flowing through the given circuit will be ______ A.
Official Ans. by NTA (2)
Sol. Equivalent circuit
I = 6/3 = 2 A

6. A closely wounded circular coil of radius 5 cm produces a magnetic field of 37.68 × 10?? T at its center. The current through the coil is ______ A. [Given, number of turns in the coil is 100 and π = 3.14]
Official Ans. by NTA (3)
Sol. B_centre = Nμ?I/(2R)
37.68 × 10?? = (100 × 4π × 10?? × I)/(2 × 5 × 10?²)
I = 3 A

7. Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be ______ I.
Official Ans. by NTA (24)
Sol. I_net = I? + I? + 2√I?√I? cos φ
I_max for φ = 0 & I_min for φ = π
I_max = (√I? + √I?)² = (√9I + √4I)² = 25I
I_min = (√I? - √I?)² = (√9I - √4I)² = I
I_max - I_min = 25I - I = 24I

8. A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is ______ A·m². [Given π = 3.14]
Official Ans. by NTA (11)
Sol. 314/100 = 2πR ⇒ R = 0.5 m
Magnetic Moment = IA = 14 × πR² = 14 × 3.14 × 1/4 = 10.99 ≈ 11.00

9. The X-Y plane be taken as the boundary between two transparent media M? and M?. M? in Z ≥ 0 has a refractive index of √2 and M? with Z < 0 has a refractive index of √3. A ray of light travelling in M? along the direction given by the vector A = 4√3 i - 3√3 j - 5k, is incident on the plane of separation. The value of difference between the angle of incident in M? and the angle of refraction in M? will be ______ degree.
Official Ans. by NTA (15)
Sol. A = 4√3 i - 3√3 j - 5k
As incident vector A makes i angle with normal z-axis & refracted vector R makes r angle with normal z-axis with help of direction cosine
i = cos?¹(A_z/A) = cos?¹(5/√((4√3)² + (3√3)² + 5²)) = cos?¹(5/10) ⇒ i = 60°
√2 sin 60 = √3 × sin r
r = 45°
Difference between i & r = 60 - 45 = 15

10. If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6 × 10?? m, will be ______ × 10? N/C.
Official Ans. by NTA (1)
Sol. E = V/d = Potential barrier Across Junction / width of Depletion layer
= 0.6 V / (6 × 10?? m) = 1 × 10? V/m = 1 × 10? N/C

CHEMISTRY

1. Which of the following pair of molecules contain odd electron molecule and an expanded octet molecule?
(A) BCl? and SF?
(B) NO and H?SO?
(C) SF? and H?SO?
(D) BCl? and NO
Official Ans. by NTA (B)
Sol. (A) BCl? → Even Electron molecule, SF? → Expanded octet molecule
(B) NO → Odd Electron molecule, H?SO? → Expanded octet.
(C) SF? → Even Electron molecule, H?SO? → Expanded octet.
(D) BCl? → Even Electron molecule, NO → Odd Electron molecule
S → 12e? in outer orbit.

2. N?(g) + 3H?(g) ? 2NH?(g)
20g 5g
Consider the above reaction, the limiting reagent of the reaction and number of moles of NH? formed respectively are:
(A) H?, 1.42 moles
(B) H?, 0.71 moles
(C) N?, 1.42 moles
(D) N?, 0.71 moles
Official Ans. by NTA (C)
Sol. N?(g) + 3H?(g) ? 2NH?(g)
W = 20g 5g
n = 20/28 5/2
Stoichiometric Amount:
N? → (20/28)/1 = 20/28, H? → (5/2)/3 = 5/6
N? is the Limiting Reagent.
n(NH?) = 2 × n(N?) = 2 × 20/28 = 1.42

TEST PAPER WITH SOLUTION

3. 100 mL of 5% (w/v) solution of NaCl in water was prepared in 250 mL beaker. Albumin from the egg was poured into NaCl solution and stirred well. This resulted in a/an :
(A) Lyophilic sol
(B) Lyophobic sol
(C) Emulsion
(D) Precipitate
Official Ans. by NTA (A)
Sol. Standard method for the preparation of lyophilic sol. (Discussed in lab Manual)

4. The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 kJmol?¹. The first ionization enthalpy (kJ mol-1) of Al is:
(A) 487
(B) 768
(C) 577
(D) 856
Official Ans. by NTA (C)
Sol. I.E.: Na < Al < Mg < Si. 496 < IE(Al) < 737
Option (C), matches the condition. i.e IE(Al) = 577 kJmol?¹

5. In metallurgy the term "gangue" is used for:
(A) Contamination of undesired earthy materials.
(B) Contamination of metals, other than desired metal
(C) Minerals which are naturally occuring in pure form
(D) Magnetic impurities in an ore.
Official Ans. by NTA (A)
Sol. Earthy and undesired materials present in the ore, other then the desired metal, is known as gangue.

6. The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives :
(A) Zn(OH)?
(B) ZnO
(C) [Zn(OH)?]²?
(D) [ZnO?]²?
Official Ans. by NTA (D)
Sol. Zinc dissolves in excess of aqueous alkali
Zn + 2OH? + 2H?O → [Zn(OH)?]²? + H?↑
Tetrahydroxozincate(II) ion
However, this reaction in NCERT is given as
Zn + 2NaOH → Na?ZnO? + H?↑
ZnO?²? is anhydrous form of [Zn(OH)?]²?
So in aqueous medium best answer of this question is [Zn(OH)?]²?

7. Lithium nitrate and sodium nitrate, when heated separately, respectively, give :
(A) LiNO? and NaNO?
(B) Li?O and Na?O
(C) Li?O and NaNO?
(D) LiNO? and Na?O
Official Ans. by NTA (C)
Sol. Li?O, NaNO?
As per NCERT Lithium nitrate when heated gives lithium oxide, Li?O, whereas other alkali metal nitrates decompose to give the corresponding nitrite.
4LiNO? → 2Li?O + 4NO? + O?
2NaNO? → 2NaNO? + O?
However, the decomposition product of NaNO? are temperature dependent process as shown in the below reaction.
NaNO? → Δ → NaNO?(s) + 1/2 O?(g)
→ Δ → Na?O(s) + N?(g) + O?(g)
As temperature is not mentioned, we can go by Ans. (C)

8. Number of lone pairs of electrons in the central atom of SCl?, O?, ClF? and SF? respectively, are :
(A) 0,1,2 and 2
(B) 2,1,2 and 0
(C) 1,2,2 and 0
(D) 2,1,2 and 0
Official Ans. by NTA (B)
Sol. SCl? (2 ?.p.), O? (1 ?.p.), ClF? (2 ?.p.), SF? (0 ?.p.)

9. In following pairs, the one in which both transition metal ions are colourless is :
(A) Sc³?, Zn²?
(B) Ti??, Cu²?
(C) V²?, Ti³?
(D) Zn²?, Mn²?
Official Ans. by NTA (A)
Sol. (A) Sc³?, Zn²?
(B) Ti??, Cu²?
3d?, 3d¹?, 3d?, 3d?
(C) V²?, Ti³?
(D) Zn²?, Mn²?
3d³, 3d¹, 3d¹?, 3d?
No d-d transitions in ions with d?, d¹? configuration. Therefore they are colourless.

10. In neutral or faintly alkaline medium, KMnO? being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :
(A) 5
(B) 1
(C) 0
(D) 3
Official Ans. by NTA (D)
Sol. 8MnO?? + 3S?O?²? + H?O → 8MnO? + 6SO?²? + 2OH?
Change in oxidation state of Mn is from +7 to +4 which is 3.

11. Which among the following pairs has only herbicides?
(A) Aldrin and Dieldrin
(B) Sodium chlorate and Aldrin
(C) Sodium arsinate and Dieldrin
(D) Sodium chlorate and sodium arsinate.
Official Ans. by NTA (D)
Sol. Both sodium chlorate and sodium arsenite behave as herbicide.

12. Which among the following is the strongest Bronsted base ?
Official Ans. by NTA (D)
Sol. It is most basic because there is no amine inversion.

13. Which among the following pairs of the structures will give different products on ozonolysis? (Consider the double bonds in the structures are rigid and not delocalized.)
Official Ans. by NTA (C)
Sol.

14. (Major Product)
Official Ans. by NTA (C)
Sol.

15. Consider the above reaction sequence, the Product 'C' is :
Official Ans. by NTA (D)
Sol.

16. 'A' (C?H?Cl?O) → C?H?ClNO → Br? → NaOH → H?N
Consider the above reaction, the compound 'A' is :
Official Ans. by NTA (C)
Sol.

17. Which among the following represent reagent 'A'?
Official Ans. by NTA (A)
Sol.

18. The product 'B' is :
Official Ans. by NTA (B)
Sol. Cross aldol condensation

19. Which of the following compounds is an example of hypnotic drug ?
(A) Seldane
(B) Amytal
(C) Aspartame
(D) Prontosil
Official Ans. by NTA (B)
Sol. Amytal is hypnotic drug used to treat sleeping disorder.

20. A compound 'X' is acidic and it is soluble in NaOH solution, but insoluble in NaHCO? solution. Compound 'X' also gives violet colour with neutral FeCl? solution. The compound 'X' is :
Official Ans. by NTA (B)
Sol. 6 C?H?OH + FeCl? → [Fe(C?H?O)?]³? violet colour

SECTION-B

1. Resistance of a conductivity cell (cell constant 129 m?¹) filled with 74.5 ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. Λ?/Λ? = x × 10?³. The value of x is ______. (Nearest integer)
Given, molar mass of KCl is 74.5 gmol?¹
Official Ans. by NTA (1000)
Sol. ?/A = 129 m?¹
KCl solution 1: 74.5 ppm, R? = 100 Ω
KCl solution 2: 149 ppm, R? = 50 Ω
Here, ppm?/ppm? = M?/M?
Λ?/Λ? = (κ? × 1000/M?)/(κ? × 1000/M?) = (κ?/κ?) × (M?/M?) = (50/100) × 2 = 1 = 1000 × 10?³
Ans. 1000

2. Ionic radii of cation A? and anion B? are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for B?. A? is present in all octahedral voids. The edge length of the unit cell of the crystal AB is ______ pm. (Nearest Integer)
Official Ans. by NTA (512)
Sol. a = 2(r? + r?)
a = 2(102 + 181)
a = 2(283)
a = 566 pm

3. The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a? (Where a? = Bohr radius 52.9 pm) is ______ km s?¹. (Given : Mass of electron = 9.1 × 10?³¹ kg, Planck's constant h = 6.63 × 10?³? Js)
Official Ans. by NTA (548)
Sol. Heisenberg's uncertainty principle
Δx × Δp_x ≥ h/4π
⇒ 2a? × mΔv_x = h/4π (minimum)
⇒ Δv_x = h/4π × 1/(2a?) × 1/m
= 6.63 × 10?³? / (4 × 3.14 × 2 × 52.9 × 10?¹² × 9.1 × 10?³¹)
= 548273 ms?¹
= 548.273 km s?¹
= 548 km s?¹

4. When 600 mL of 0.2 M HNO? is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ______ × 10?² °C.
(Enthalpy of neutralisation = 57 kJ mol?¹ and Specific heat of water = 4.2 J K?¹ g?¹)
(Neglect heat capacity of flask)
Official Ans. by NTA (54)
Sol. HNO?: 600 mL × 0.2 M = 120 m mol
NaOH: 400 mL × 0.1 M = 40 m mol
HNO? + NaOH → NaNO? + H?O
Bef. 120 40
Aft. 80 0 40 m mol
Δ_rH = 40 m mol × (57 × 10³) J/mol = 40 × 10?³ mol × 57 × 10³ J/mol = 2280 J
mSΔT = 2280
⇒ 1000 mL × 1 g/mL × 4.2 × ΔT = 2280
ΔT = 2280/4.2 × 10?³ = 22800/42 × 10?³ = 542.86 × 10?³ = 54.286 × 10?² K
Ans. 54 (Closest integer)

5. If O? gas is bubbled through water at 303 K, the number of millimoles of O? gas that dissolve in 1 litre of water is ______. (Nearest Integer)
(Given : Henry's Law constant for O? at 303 K is 46.82 kbar and partial pressure of O? = 0.920 bar)
(Assume solubility of O? in water is too small, nearly negligible)
Official Ans. by NTA (1)
Sol. p = K_H × x
0.920 = 46.82 × 10³ bar × (mol of O?)/(mol of H?O)
0.920 = 46.82 × n_O?
n_O? = 0.920/(46.82 × 18) = 1.09 × 10?³
⇒ m mol of O? = 1

MATHEMATICS

1. Let R be a relation from the set {1, 2, 3, 60} to itself such that R = {(a,b) : b = pq, where p,q ≥ 3 are prime numbers}. Then, the number of elements in R is:
(A) 600
(B) 660
(C) 540
(D) 720
Official Ans. by NTA (B)
Sol. Number of possible values of a = 60 for b = pq
If p = 3, q = 3,5,7,11,13,17,19
If p = 5, q = 5,7,11
If p = 7, q = 7
Total cases = 60 × 11 = 660

2. If z = 2 + 3i, then z? + (z?)? is equal to :
(A) 244
(B) 224
(C) 245
(D) 265
Official Ans. by NTA (A)
Sol. z? + (z?)? = (2 + 3i)? + (2 - 3i)?
= 2(?C?2? + ?C?2³(3i)² + ?C?2¹(3i)?)
= 2(32 + 10 × 8(-9) + 5 × 2 × 81) = 244

3. Let A and B be two 3×3 non-zero real matrices such that AB is a zero matrix. Then
(A) The system of linear equations AX = 0 has a unique solution
(B) The system of linear equations AX = 0 has infinitely many solutions
(C) B is an invertible matrix
(D) adj (A) is an invertible matrix
Official Ans. by NTA (B)
Sol. AB = 0 ⇒ |AB| = 0
If |A| ≠ 0, B = 0 (not possible)
If |B| ≠ 0, A = 0 (not possible)
Hence |A| = |B| = 0
⇒ AX = 0 has infinitely many solutions

4. If 1/((20 - a)(40 - a)) + 1/((40 - a)(60 - a)) + ... + 1/((180 - a)(200 - a)) = 1/256 then the maximum value of a is :
(A) 198
(B) 202
(C) 212
(D) 218
Official Ans. by NTA (C)
Sol. By splitting
1/20 [(1/(20 - a) - 1/(40 - a)) + (1/(40 - a) - 1/(60 - a)) + ... + (1/(180 - a) - 1/(200 - a))] = 1/256
⇒ 1/20 (1/(20 - a) - 1/(200 - a)) = 1/256
(20 - a)(200 - a) = 256 × 9
a² - 220a + 1696 = 0
a = 8, 212
Hence maximum value of a is 212.

5. If lim_{x→0} (αe^x + βe^{-x} + γ sin x)/(x sin²x) = 2/3, where α, β, γ ∈ R, then which of the following is NOT correct?
(A) α² + β² + γ² = 6
(B) αβ + βγ + γα + 1 = 0
(C) αβ² + βγ² + γα² + 3 = 0
(D) α² - β² + γ² = 4
Official Ans. by NTA (C)
Sol. constant terms should be zero
⇒ α + β = 0
coeff of x should be zero
⇒ α - β + γ = 0
coeff of x² should be zero
lim_{x→0} (x³(α/3! - β/3! - γ/3!) + x?(α/4! - β/4! - γ/4!))/x³ = 2/3
⇒ α/6 - β/6 - γ/6 = 2/3
⇒ α = 1, β = -1, γ = -2

6. The integral ∫?^{π/2} 1/(3 + 2sin x + cos x) dx is equal to:
(A) tan?¹(2)
(B) tan?¹(2) - π/4
(C) 1/2 tan?¹(2) - π/8
(D) 1/2
Official Ans. by NTA (B)
Sol. I = ∫?^{π/2} dx/(3 + 2sin x + cos x) = ∫?^{π/2} sec²(x/2) dx / (2tan²(x/2) + 4tan(x/2) + 4)
Put tan(x/2) = t, so
I = ∫?¹ dt/((t + 1)² + 1) = tan?¹(x + 1)|?¹ = tan?¹2 - π/4

7. Let the solution curve y = y(x) of the differential equation (1 + e^{2x})(dy/dx + y) = 1 pass through the point (0, π/2). Then, lim_{x→∞} e^x y(x) is equal to :
(A) π/4
(B) 3π/4
(C) π/2
(D) 3π/2
Official Ans. by NTA (B)
Sol. dy/dx + y = 1/(1 + e^{2x})
So integrating factor is e^{∫1 dx} = e^x
So solution is y·e^x = tan?¹(e^x) + c
Now as curve is passing through (0, π/2) so
⇒ c = π/4
⇒ lim_{x→∞} (y·e^x) = lim_{x→∞} (tan?¹(e^x) + π/4) = 3π/4

8. Let a line L pass through the point of intersection of the lines bx + 10y - 8 = 0 and 2x - 3y = 0, b ∈ R - {4/3}. If the line L also passes through the point (1, 1) and touches the circle 17(x² + y²) = 16 then the eccentricity of the ellipse x²/5 + y²/5 = 1 is :
(A) 2/√5
(B) √(3/5)
(C) 1/√5
(D) √(2/5)
Official Ans. by NTA (B)
Sol. Line is passing through intersection of bx + 10y - 8 = 0 and 2x - 3y = 0 is (bx + 10y - 8) + λ(2x - 3y) = 0. As line is passing through (1,1) so λ = b + 2

9. If the foot of the perpendicular from the point A(-1,4,3) on the plane P: 2x + my + nz = 4 is (-2, 7/2, 3/2), then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3, -1, -4, is equal to :
(A) 1
(B) √26
(C) 2√2
(D) √14
Official Ans. by NTA (B)
Sol. Let B be foot of ⊥ coordinates of B = (-2, 7/2, 3/2)
Direction ratio of line AB is <2, 1, 3> so m = 1, n = 3
So equation of AC is (x + 1)/3 = (y - 4)/(-1) = (z - 3)/(-4) = λ
So point C is (3λ - 1, -λ + 4, -4λ + 3). But C lies on the plane, so
6λ - 2 - λ + 4 - 12λ + 9 = 4
⇒ λ = 1 ⇒ C(2, 3, -1)
⇒ AC = √26

10. Let a = 3i + j and b = i + 2j + k. Let c be a vector satisfying a × (b × c) = b + λc. If b and c are non-parallel, then the value of λ is :
(A) -5
(B) 5
(C) 1
(D) -1
Official Ans. by NTA (A)
Sol. a = 3i + j, b = i + 2j + k
a × (b × c) = b + λc
⇒ (a·c)b - (a·b)c = b + λc
⇒ a·c = 1, a·b = -λ
⇒ (3i + j)·(i + 2j + k) = -λ
⇒ λ = -5

11. The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos?¹(3/√13). If the distance of the point B from the tower is 15 units, then cot α is equal to :
(A) 6/5
(B) 9/5
(C) 4/3
(D) 7/3
Official Ans. by NTA (A)
Sol. given OB = 15, cos β = 3/√13, tan β = 2/3
tan β = h/15 ⇒ 2/3 = h/15 ⇒ h = 10
OA² + AB² = 225 ⇒ OA² + 81 = 225 ⇒ OA = 12
tan α = 10/12 ⇒ cot α = 12/10 = 6/5

12. The statement (p ∧ q) ⇒ (p ∧ r) is equivalent to :
(A) q ⇒ (p ∧ r)
(B) p ⇒ (p ∧ r)
(C) (p ∧ r) ⇒ (p ∧ q)
(D) (p ∧ q) ⇒ r
Official Ans. by NTA (D)
Sol. (p ∧ q) ⇒ (p ∧ r)
~ (p ∧ q) ∨ (p ∧ r)
~ (~p ∨ ~q) ∨ (p ∧ r)
~ (~p ∨ (p ∧ r)) ∨ ~q
~ (~p ∨ p) ∧ (~p ∨ r) ∨ ~q
~ (~p ∨ r) ∨ ~q
~ (~p ∨ ~q) ∨ r
~ (p ∧ q) ∨ r
(p ∧ q) ⇒ r

13. Let the circumference of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k?, k?), then k? + k? is equal to :
(A) 2
(B) 4/7
(C) 2/7
(D) 4
Official Ans. by NTA (B)
Sol. m_AC → ∞
m_PD = 0
D((a + a)/2, (b + 3)/2)
D(a, (b + 3)/2)
m_PD = 0
b + 3 - 2 = 0
b = -1
E((b + a)/2, (5 + b)/2) = (a/2, 2)
m_CB · m_EP = -1
(6/(-1 - a)) = (2/(a - 3)) = -1
12 = (1 + a)(a - 3)
12 = a² - 3a + a - 3
⇒ a² - 2a - 15 = 0
(a - 5)(a + 3) = 0

14. Let â and b? be two unit vectors such that the angle between them is π/4. If θ is the angle between the vectors (â + b?) and (â + 2b? + 2(â × b?)), then the value of 164 cos²θ is equal to :
(A) 90 + 27√2
(B) 45 + 18√2
(C) 90 + 3√2
(D) 54 + 90√2
Official Ans. by NTA (A)
Sol. â∧b? = π/4 = φ
â·b? = |â||b?|cos φ = cos φ = 1/√2
cos θ = ((â + b?)·(â + 2b? + 2(â × b?)))/(|â + b?||â + 2b? + 2(â × b?)|)
|â + b?|² = 2 + 2â·b? = 2 + √2
â × b? = (1/√2) n?
|â + 2b? + 2c|² = 1 + 4 + 4/2 + 4â·b? + 8b?·c + 4c·â = 7 + 4/√2 = 7 + 2√2
(â + b?)·(â + 2b? + 2c) = 1 + 2/√2 + 1/√2 + 2 = 3 + 3/√2
cos θ = (3 + 3/√2)/(√(2 + √2)√(7 + 2√2))
cos² θ = 9(√2 + 1)² / (2(2 + √2)(7 + 2√2))
164 cos² θ = 90 + 27√2

15. If f(α) = ∫?^α log??t/(1 + t) dt, α > 0, then f(e³) + f(e?³) is equal to :
(A) 9
(B) 9/2
(C) 9/log_e(10)
(D) 9/(2log_e(10))
Official Ans. by NTA (D)
Sol. f(e³) = ∫?^{e³} ln t/((ln 10)(1 + t)) dt ...(1)
f(α) = ∫?^α ln t/((ln 10)(1 + t)) dt
t = 1/x ⇒ x = 1/t
dt = -1/x² dx
f(α) = ∫?^{1/α} -ln x/((ln 10)(1 + 1/x)) (-1/x²) dx = 1/ln10 ∫?^{1/α} ln x/(x(x + 1)) dx
f(e?³) = 1/ln10 ∫?^{e³} ln t/(t(t + 1)) dt ...(2)
Add (1) & (2)
f(e³) + f(e?³) = (1/ln10) ∫?^{e³} ln t/(1 + t) [1 + 1/t] dt
= (1/ln10) ∫?^{e³} ln t/t dt
ln t = r
dt/t = dr
= (1/ln10) ∫?³ r dr = (1/ln10)(r²/2)|?³ = (1/log_e 10)(9/2) = 9/(2log_e 10)

16. The area of the region {(x,y) : |x - 1| ≤ y ≤ √(5 - x²)} is equal to :
(A) 5/2 sin?¹(3/5) - 1/2
(B) 5π/4 - 3/2
(C) 3π/4 + 3/2
(D) 5π/4 - 1/2
Official Ans. by NTA (D)
Sol. |x - 1| < y < √(5 - x²)
When |x - 1| = √(5 - x²)
⇒ (x - 1)² = 5 - x²
⇒ x² - x - 2 = 0
⇒ x = 2, -1
Required Area = Area of ΔABC + Area of region BCD
= 1/2 |1 0 1; 2 1 1; -1 2 1| + π/4(√5)² - 1/2(√5)²
= 5π/4 - 1/2

17. Let the focal chord of the parabola P: y² = 4x along the line L: y = mx + c, m > 0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H: x² - y² = 4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :
(A) 2√6
(B) 2√14
(C) 4√6
(D) 4√14
Official Ans. by NTA (B)
Sol. H: x²/4 - y²/4 = 1
Focus (ae, 0) = F(2√2, 0)
Line L: y = mx + c pass (1,0)
0 = m + c ...(1)
Line L is tangent to Hyperbola. x²/4 - y²/4 = 1
C = ±√(a²m² - ?²)
C = ±√(4m² - 4)
From (1)
-m = ±√(4m² - 4)
Squaring
m² = 4m² - 4
4 = 3m²
|2/√3| = m (as m > 0)
C = -m
C = -2/√3

18. The number of points, where the function f: R → R f(x) = |x - 1| cos |x - 2| sin |x - 1| + (x - 3)|x² - 5x + 4| is NOT differentiable, is :
(A) 1
(B) 2
(C) 3
(D) 4
Official Ans. by NTA (B)
Sol. f(x) = |x - 1| cos |x - 2| sin |x - 1| + (x - 3)|x² - 5x + 4|
= |x - 1| cos |x - 2| sin |x - 1| + (x - 3)|x - 1||x - 4|
= |x - 1| [cos |x - 2| sin |x - 1| + (x - 3)|x - 4|]
Non differentiable at x = 1 and x = 4

19. Let S = {1,2,3,...,2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022) = 1, is :
(A) 128/1011
(B) 166/1011
(C) 127/337
(D) 112/337
Official Ans. by NTA (D)
Sol. Total number of elements = 2022
2022 = 2 × 3 × 337
HCF (n, 2022) = 1 is feasible when the value of 'n' and 2022 has no common factor.
A = Number which are divisible by 2 from {1,2,3....2022}
n(A) = 1011
B = Number which are divisible by 3 from {1,2,3....2022}
n(B) = 674
A∩B = Number which are divisible by 6 from {1,2,3....2022}
6,12,18,...,2022
337 = n(A∩B)
n(A∪B) = n(A) + n(B) - n(A∩B) = 1011 + 674 - 337 = 1348
C = Number which divisible by 337 from {1,....2022}
C = {337,674,1011,1348,1685,2022}
Already counted in Set (A∪B): 674, 1348
Already counted in Set (A∪B): 2022
Total elements which are divisible by 2 or 3 or 337 = 1348 + 2 = 1350
Favorable cases = Element which are neither divisible by 2, 3 or 337 = 2022 - 1350 = 672
Required probability = 672/2022 = 112/337

20. Let f(x) = 3^{(x² - 2)³ + 4}, x ∈ R. Then which of the following statements are true?
P : x = 0 is a point of local minima of f
Q : x = √2 is a point of inflection of f
R : f is increasing for x > √2
(A) Only P and Q
(B) Only P and R
(C) Only Q and R
(D) All, P, Q and R
Official Ans. by NTA (D)
Sol. f(x) = 81·3^{(x² - 2)³}
f'(x) = 81·3^{(x² - 2)³} ln3 · 3(x² - 2)² · 2x
= (81 × 6)3^{(x² - 2)³} x(x² - 2)² ln3
x = 0 is point of local min
f''(x) = (486 ln3)3^{(x² - 2)³} x(x² - 2)²
g'(x) = 3^{(x² - 2)³}(x² - 2)² + x·3^{(x² - 2)³}·4x·(x² - 2) + x·(x² - 2)²·3^{(x² - 2)³} ln3·(x² - 2)²·2x
= 3^{(x² - 2)³}(x² - 2)[x² - 2 + 4x² + 6x² ln3(x² - 2)³]
g'(x) = 3^{(x² - 2)³}(x² - 2)[5x² - 2 + 6x² ln3(x² - 2)³]
f''(x) = k g'(x)
f''(√2) = 0, f''(√2?) > 0, f''(√2?) < 0
x = √2 is point of inflection
f''(x) > 0 for x > √2 so f(x) is increasing

SECTION-B

1. Let S = {θ ∈ (0,2π) : 7cos²θ - 3sin²θ - 2cos²2θ = 2}. Then, the sum of roots of all the equations x² - 2(tan²θ + cot²θ)x + 6sin²θ = 0, θ ∈ S, is
Official Ans. by NTA (16)
Sol. 7cos²θ - 3sin²θ - 2cos²2θ = 2
4cos²θ + 3cos2θ - 2cos²2θ = 2
2(1 + cos2θ) + 3cos2θ - 2cos²2θ = 2
2cos²2θ - 5cos2θ = 0
cos2θ(2cos2θ - 5) = 0
cos2θ = 0

2. Let the mean and the variance of 20 observations x?,x?,...,x?? be 15 and 9, respectively. For α ∈ R, if the mean of (x? + α)², (x? + α)², ..., (x?? + α)² is 178, then the square of the maximum value of α is equal to
Official Ans. by NTA (4)
Sol. Σx? = 15 × 20 = 300 ...(i)
Σx?²/20 - (15)² = 9 ...(ii)
Σx?² = 234 × 20 = 4680
Σ(x? + α)²/20 = 178 ⇒ Σ(x? + α)² = 3560
⇒ Σx?² + 2αΣx? + Σα² = 3560
4680 + 600α + 20α² = 3560
⇒ α² + 30α + 56 = 0
⇒ (α + 28)(α + 2) = 0
α = -2, -28
Square of maximum value of α is 4

3. Let a line with direction ratios a, -4a, -7 be perpendicular to the lines with direction ratios 3, -1, 2b and b, a, -2. If the point of intersection of the line (x + 1)/(a² + b²) = (y - 2)/(a² - b²) = z/1 and the plane x - y + z = 0 is (α, β, γ), then α + β + γ is equal to
Official Ans. by NTA (10)
Sol. (a, -4a, -7) ⊥ to (3, -1, 2b) ⇒ a = 2b ...(i)
(a, -4a, -7) ⊥ to (b, a, -2) ⇒ 3a + 4a - 14b = 0 ⇒ ab - 4a² + 14 = 0 ...(ii)
From Equations (i) and (ii)
2b² - 16b² + 14 = 0 ⇒ b² = 1 ⇒ a² = 4b² = 4
(x + 1)/5 = (y - 2)/3 = z/1 = k
α = 5k - 1, β = 3k + 2, γ = k
As (α, β, γ) satisfies x - y + z = 0
5k - 1 - (3k + 2) + k = 0 ⇒ k = 1
α + β + γ = 9k + 1 = 10

4. Let a?, a?, a?, ... be an A.P. If Σ_{r=1}^∞ a_r/2^r = 4, then 4a? is equal to
Official Ans. by NTA (16)
Sol. S = a?/2 + a?/2² + a?/2³ + ...
S/2 = a?/2² + a?/2³ + ...
S/2 = a?/2 + d(1/2² + 1/2³ + ...)
S/2 = a?/2 + d(1/(1 - 1/2))
∴ S = a? + d = a? = 4
Or 4a? = 16

5. Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (?√2 + 1/³√3)? in the increasing powers of 1/³√3 be ?√6 : 1. If the sixth term from the beginning is α/³√3, then α is equal to
Official Ans. by NTA (84)
Sol. T?/T_{n-3} = (?C?(2^{1/4})^{n-4}(3^{-1/4})?)/(?C_{n-4}(2^{1/4})?(3^{-1/4})^{n-4}) = ?√6/1
⇒ 2^{(n-8)/4}3^{(n-8)/4} = 6^{1/4}
⇒ 6^{n-8} = 6
⇒ n - 8 = 1 ⇒ n = 9
T? = ?C?(2^{1/4})?(3^{-1/4})? = 84/?√3
∴ α = 84

6. The number of matrices of order 3×3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is
Official Ans. by NTA (282)
Sol. A = [a_{ij}]_{3×3}, a_{ij} ∈ {0,1}
Σa_{ij} = 2,3,5,7
Total matrix = ?C? + ?C? + ?C? + ?C? = 282

7. Let p and p + 2 be prime numbers and let Δ = |p! (p+1)! (p+2)!; (p+1)! (p+2)! (p+3)!; (p+2)! (p+3)! (p+4)!|. Then the sum of the maximum values of α and β such that p^α and (p+2)^β divide Δ, is
Official Ans. by NTA (4)
Sol. Δ = |P! (P+1)! (P+2)!; (P+1)! (P+2)! (P+3)!; (P+2)! (P+3)! (P+4)!|
Δ = P!(P+1)!(P+2)! |1 P+1 (P+2)(P+1); 1 P+2 (P+3)(P+2); 1 P+3 (P+4)(P+3)|
Δ = 2P!(P+1)!(P+2)!
Which is divisible by P^α & (P+2)^β
∴ α = 3, β = 1
Ans. 4

8. If 1/(2×3×4) + 1/(3×4×5) + 1/(4×5×6) + ... + 1/(100×101×102) = k/101, then 34k is equal to
Official Ans. by NTA (286)
Sol. 1/(2·3·4) + 1/(3·4·5) + ... + 1/(100·101·102) = k/101
(4-2)/(2·3·4) + (5-3)/(3·4·5) + ... + (102-100)/(100·101·102) = 2k/101
(1/(2·3) - 1/(3·4)) + (1/(3·4) - 1/(4·5)) + ... + (1/(100·101) - 1/(101·102)) = 2k/101
1/(2·3) - 1/(101·102) = 2k/101
∴ 2k = 101/6 - 1/102
∴ 34k = 286

9. Let S = {4,6,9} and T = {9,10,11,...,1000}. If A = {a? + a? + ... + a_k : k ∈ N, a?,a?,...,a_k ∈ S} then the sum of all the elements in the set T - A is equal to

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