24th Feb. 2021 | Shift - 2
PHYSICS

1. Zener breakdown occurs in a p-n junction having p and n both :
(1) lightly doped and have wide depletion layer.
(2) heavily doped and have narrow depletion layer.
(3) heavily doped and have wide depletion layer.
(4) lightly doped and have narrow depletion layer.
Ans. (2)
Sol. The zener breakdown occurs in the heavily doped p-n junction diode. Heavily doped p-n junction diodes have narrow depletion region.

2. According to Bohr atom model, in which of the following transitions will the frequency be maximum?
(1) n = 2 to n = 1
(2) n = 4 to n = 3
(3) n = 5 to n = 4
(4) n = 3 to n = 2
Ans. (1)
Sol. 5 4 3 2 ΔE = hf
f is more for transition from n = 2 to n = 1

3. An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
(1) 10?² nm
(2) 10?³ nm
(3) 10?? nm
(4) 10?¹ nm
Ans. (2)
Sol. λ_min = hc/eV
λ_min = (1240 nm - eV)/(1.24 × 10?)
λ_min = 10?³ nm

4. On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:
(1) suffer change in momentum when impinge on the walls of container.
(2) continuously stick to the walls of container.
(3) continuously lose their energy till it reaches wall.
(4) are attracted by the walls of container.
Ans. (1)
Sol. On the basis of kinetic theory of gases, the gas pressure is due to the molecules suffering change in momentum when impinge on the walls of container.

5. 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 :
(1) 10/11 a
(2) 2/3 a
(3) 1/6 a
(4) 5/6 a
Ans. (4)
Sol. Let σ is the surface mass density of disc.
X_com = (σ × πa² × a - σ (πa²/4) × (3a/2)) / (σπa² - σπa²/4)
X_com = (a - 3a/8) / (1 - 1/4)
X_com = 5a/3
X_com = 5a/6

6. 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.
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false.
(4) Both Statement I and Statement II are false
Ans. (1)
Sol. 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

7. When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :
(1) elliptical
(2) parabolic
(3) straight line
(4) circular
Ans. (1)
Sol. We know that is SHM;
V = ω√(A² - x²)
elliptical

8. Match List - I with List - II.
List - I
(a) Source of microwave frequency
(b) Source of infrared frequency
(c) Source of Gamma Rays
(d) Source of X-rays
List - II
(i) Radioactive decay of nucleus
(ii) Magnetron
(iii) Inner shell electrons
(iv) Vibration of atoms and molecules
(v) LASER
(vi) RC circuit
Choose the correct answer from the options given below :
(1) (a)- (ii), (b)- (iv), (c)- (i), (d)- (iii)
(2) (a)- (vi), (b)- (iv), (c)- (i), (d)- (v)
(3) (a)- (ii), (b)- (iv), (c)- (vi), (d)- (iii)
(4) (a)- (vi), (b)- (v), (c)- (i), (d)- (iv)
Ans. (1)
Sol.
(a) Source of microwave frequency - (ii) Magnetron
(b) Source of infra red frequency - (iv) Vibration of atom and molecules
(c) Source of gamma ray - (i) Radio active decay of nucleus
(d) Source of X-ray - (iii) inner shell electron

9. A O C B The logic circuit shown above is equivalent to : A O (1) B O (2) A O (3) B O (4) B O Ans. (2) Sol. A O B O C = A + B C = AB

10. If the source of light used in a Young's double slit experiment is changed from red to violet:
(1) the fringes will become brighter.
(2) consecutive fringe lines will come closer.
(3) the central bright fringe will become a dark fringe.
(4) the intensity of minima will increase.
Ans. (2)
Sol. β = λD/d
As λ_v < λ_R
⇒ β_v < β_R
: Consecutive fringe line will come closer.
: (2)

11. 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.]
(1) 49 N
(2) 49.83 N
(3) 49.17 N
(4) 48.83 N
Ans. (4)
Sol. At north pole, weight
Mg = 49
Now, at equator g' = g - ω²R
⇒ Mg' = M(g - ω²R)
⇒ weight will be less than Mg at equator.

12. If one mole of an ideal gas at (P?, V?) 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 workdone by the gas is equal to :
Ans. (3)
Sol. AB → Isothermal process
W_AB → nRT ln 2 = RT ln 2
BC → Isochoric process
W_BC = 0
CA → Adiabatic process
W_CA = (P?V? - (P?/4) × 2V?)/(1 - γ) = P?V?/(2(1 - γ)) = RT/(2(1 - γ))
W_ABCA = RT ln 2 + RT/(2(1 - γ))
= RT[ln 2 - 1/(2(γ - 1))]

13. 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 :
(1) 1.33%
(2) 1.30%
(3) 1.13%
(4) 1.03%
Ans. (3)
Sol. T = 2π√(?/g)
T² = 4π²[?/g]
g = 4π²[?/T²]
Δg/g = Δ?/? + 2ΔT/T
= [1 mm/1 m + 2(10 × 10?³)/1.95] × 100
= 1.13%

14. 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 :
(1) 1/(2π)√(2k/(Mg sinα))
(2) 1/(2π)√(k/(Mg sinα))
(3) 1/(2π)√(2k/M)
(4) 1/(2π)√(k/(2M))
Ans. (1)
Sol. Equivalent K = K + K = 2K
Now, T = 2π√(m/K_eq)
⇒ T = 2π√(m/2k)
∴ f = 1/(2π)√(2k/m)

15. 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 :
(1) 9A
(2) 3.0 A
(3) 2.25 A
(4) 3.37 A
Ans. (3) Sol. When switch S is closed
9V Given : ν = 9ν From ν = IR
I = ν/R
R_eq. = 2 + 2 = 4Ω
I = 9/4 = 2.25A

16. The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is :
(1) 4:2
(2) 4:1
(3) 1:4
(4) 4:3
Ans. (2)
Sol. From De-broglie's wavelength :-
λ = h/mv
Given λ_p = λ_a
v α 1/m
v_p/v_a = m_a/m_p = 4m_p/m_p = 4/1

17. 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< (1) (q²/(2πε?md³))^(1/2)
(2) (πε?md³/(2q²))^(1/2)
(3) (2πε?md³/q²)^(1/2)
(4) (2q²/(πε?md³))^(1/2)
Ans. (1)
Sol. Restoring force on proton :-

18. A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains :
(1) decrease in size and changes orientation.
(2) may increase or decrease in size and change its orientation.
(3) increase in size but no change in orientation.
(4) have no relation with external magnetic field.
Ans. (2)
Sol. Atoms of ferromagnetic material in unmagnetised 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.

19. Which of the following equations represents a travelling wave?
(1) y = Ae^(-x²)(vt + 0)
(2) y = A sin(15x - 2t)
(3) y = Ae^x cos(ωt - 0)
(4) y = A sin x cos ωt
Ans. (2)
Sol. 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) → linear function in x and t.

20. A particle is projected with velocity v? 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 :
(1) (2v?/(3α))^(1/3)
(2) (3v?²/(2α))^(1/2)
(3) (3v?²/(2α))^(-1/3)
(4) (2v?²/(3α))^(1/2)
Ans. Bonus

1. 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.
Ans.
2
γ = (F/A)/(Δ?/?)
⇒ F/A = γ Δ?/?
⇒ F/A = γ × 0.04/? ...(1)
When length & diameter is doubled.
⇒ F/(4A) = γ × Δ?/(2?) ...(2)
(F/A)/(F/4A) = (γ × 0.04/?)/(γ × Δ?/(2?))
4 = 0.04 × 2/Δ?
Δ? = 0.02
Δ? = 2 × 10?²
∴ x = 2

2. A cylindrical wire of radius 0.5 mm and conductivity 5 × 10? 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 ______.
Ans. 5
Sol. We know that J = σE
⇒ J = 5 × 10? × 10 × 10?³
⇒ J = 50 × 10? A/m²
Current flowing; I = J × πR²
I = 50 × 10? × π(0.5 × 10?³)²
I = 5 × 10? × π × 0.25 × 10??
I = 125 × 10?³ π
X = 5

3. 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.]
Ans. 8
Speed = 7.2 km/h = 2 m/s Frequency as heard by A
f_A' = f?((v + v?)/(v - v_s))
f_A' = 676((340 + 2)/(340 - 2))
f_A' = 684 Hz
∴ f_Beat = f_A' - f_B
= 684 - 676
= 8 Hz

4. 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 hexagon is ______ × 10?¹ kg m².
Ans. 8
Sol.
MOI of AB about P : I_ABp = (M/6)(?/6)²/12
MOI of AB about O,
I_ABO = [(M/6)(?/6)²/12 + M/6(?√3/2)²]
I_Hexagon = 6I_ABO = M[?²/(12×36) + ?²/36 × 3/4]
= 6/100 [24×24/(12×36) + 24×24/36 × 3/4]
= 0.8 kgm²
= 8 × 10?² kg/m²

5. 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.
Ans. 226
Sol. Using Gauss law, it is a part of cube of side 12 cm and charge at centre so;
φ = Q/6ε? = 12μC/6ε? = 2 × 4π × 9 × 10? × 10??
= 226 × 10³ Nm²/C

6. 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 ______.
Ans. 2
Sol. Given that, M?/M? = 1/2
Also, p? = p? = p
⇒ M?V? = M?V? = p
Also, we know that

7. The root mean square speed of molecules of a given mass of a gas at 27°C and 1 atmosphere pressure is 200 ms?¹. The root mean square speed of molecules of the gas at 127°C and 2 atmosphere pressure is x/√3 ms?¹. The value of x will be ______.
Ans. 400 m/s
Sol. V_rms = √(3RT?/M?)
200 = √(3R × 300/M?) ...(1)
Also, x/√3 = √(3R × 400/M?) ...(2)
(1) ÷ (2)
200/(x/√3) = √(300/400) = √(3/4)
⇒ x = 400 m/s

8. A series LCR circuit is designed to resonate at an angular frequency ω? = 10? rad/s. The circuit draws 16W power from 120 V source at resonance. The value of resistance 'R' in the circuit is ______ Ω
Ans. 900
Sol. P = V²/R
16 = 120²/R ⇒ R = 14400/16
⇒ R = 900 Ω

9. 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.
Ans. 667
Sol. f = 3 GHz, ε_r = 2.25
v = λf ⇒ λ = v/f
C = 1/√(μ?ε?)
v = 1/√(μ?μ_r ε?ε_r) ⇒ λ = 1/(f_v √(μ?ε?) · √(μ_r ε_r)) f
⇒ λ = C/(f_v √(μ_r · ε_r)) ⇒ λ = (3×10?)/(3×10? × √1 × √2.25)
⇒ λ = 667 × 10?² cm

10. 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 receiver is 10?? W. The value of x is ______.
[Gain in dB = 10 log??(P?/P?)]
Ans. 8
Sol. Power of signal transmitted : P? = 0.1 kW = 100 w
Rate of attenuation = -5 dB/Km
Total length of path = 20 km
Total loss suffered = -5 × 20 = -100 dB
Gain in dB = 10 log??(P?/P?)
-100 = 10 log??(P?/P?)
⇒ log??(P?/P?) = 10
⇒ log??(P?/P?) = log??10¹?
⇒ 100/P? = 10¹?
⇒ P? = 1/10? = 10??
∴ x = 8

24th Feb. 2021 | Shift - 2
CHEMISTRY

1. The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is :
(1) (iv) < (i) < (iii) < (ii)
(2) (iv) < (i) < (ii) < (iii)
(3) (i) < (ii) < (iii) < (iv)
(4) (iv) < (iii) < (ii) < (i)
Ans. (3)
Sol.
Reactivity α -m group present at O/P position.

2. Match List-I with List-II
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options given below :
(1) (a)- (iv), (b)- (iii), (c)- (ii), (d)- (i)
(2) (a)- (i), (b)- (ii), (c)- (iii), (d)- (iv)
(3) (a)- (iii), (b)- (i), (c)- (iv), (d)- (ii)
(4) (a)- (ii), (b)- (iv), (c)- (i), (d)- (iii)
Ans. (3)
Sol. Siderite FeCO? Calamine ZnCO? Kaolinite Si?Al?O?(OH)? or Al?O?.2SiO?.2H?O Malachite CuCO?.Cu(OH)?

3. Match List-I with List-II
List-I List-II
(Salt) (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:
(1) (a)- (ii), (b)- (i), (c)- (iv), (d)- (iii)
(2) (a)- (ii), (b)- (iv), (c)- (iii), (d)- (ii)
(3) (a)- (iv), (b)- (ii), (c)- (iii), (d)- (i)
(4) (a)- (i), (b)- (iv), (c)- (ii), (d)- (iii)
Ans. (2)
Sol. Range of visible region : - 390nm - 760nm VIBGYOR Violet Red LiCl Crimson Red NaCl Golden yellow RbCl Violet CsCl Blue So Lid Which is crimson have wave length closed to red in the spectrum of visible region which is as per given data is.

4. 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
Ans. (2)
Sol. Hydrogen is most abundant element in universe because all luminous body of universe i.e. stars & nebulae are made up of hydrogen which acts as nuclear fuel & fusion reaction is responsible for their light.

5. Given below are two statements:
Statement I : The value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life.
Statement II : The optimum value of BOD is 6.5 ppm.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Ans. (3)
Sol. For survival of aquatic life dissolved oxygen is responsible its optimum limit 6.5 ppm and optimum limit of BOD ranges from 10-20 ppm & BOD stands for biochemical oxygen demand.

6. Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of HgSO? and H?SO? ?
(1) CH? - CH? - C - H
(2)
(3) CH? - C - H
(4) CH? - C - CH?CH?
Ans. (1)
Sol. Reaction of Alkyne with HgSO? & H?SO? follow as
CH≡CH → HgSO?, H?SO? / H?O → CH?CHO
CH? - C≡CH → HgSO?, H?SO? / H?O → CH? - C - CH?
Hence, by this process preparation of CH?CH?CHO Can't possible.

7. Which one of the following compounds is non-aromatic ?
(1)
(2)
(3)
(4)
Ans. (2)
Sol.
Hence It is non-aromatic.

8. The incorrect statement among the following is :
(1) VOSO? is a reducing agent
(2) Red colour of ruby is due to the presence of CO³?
(3) Cr?O? is an amphoteric oxide
(4) RuO? is an oxidizing agent
Ans. (2)
Sol. Red colour of ruby is due to presence of CrO? or Cr?? not CO³?

9. 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/n?
Choose the most appropriate answer from the options given below:
(1) (C) only
(2) (A) and (D) only
(3) (A) only
(4) (A), (C) and (D) only
Ans. (2) Correction on NTA
Sol. (A) KE = -TE = 13.6 × Z²/n² eV
KE ∝ Z²/n²
(B) V = 2.188 × 10? × Z/n m/sec. So, Vn ∝ Z
(C) Frequency = V/(2πr) So, F ∝ Z²/n³ [? r ∝ n²/Z and v ∝ Z/n]
(D) Force ∝ Z/r² So, F ∝ Z³/n?
So, only statement (A) is correct

10. Match List-I with List-II
List-I List-II
(a) Valium (i) Antifertility drug
(b) Morphine (ii) Pernicious anaemia
(c) Norethindrone (iii) Analgesic
(d) Vitamin B?? (iv) Tranquilizer
(1) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(2) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Ans. (4)
Sol. (a) Valium (iv) Tranquilizer
(b) Morphine (iii) Analgesic
(c) Norethindrone (i) Antifertility drug
(d) Vitamin B?? (ii) Pernicious anaemia

11. The Correct set from the following in which both pairs are in correct order of melting point is :
(1) LiF > LiCl ; NaCl > MgO
(2) LiF > LiCl ; MgO > NaCl
(3) LiCl > LiF ; NaCl > MgO
(4) LiCl > LiF ; MgO > NaCl
Ans. (2)
Sol. Generally M.P. < Lattice energy = KQ1Q2/r+ r- (packing efficiency)

12. The calculated magnetic moments (spin only value) for species [FeCl] [Co(C?O?)] and MnO?²? respectively are :
(1) 5.92, 4.90 and 0 BM
(2) 5.82, O and 0 BM
(3) 4.90, 0 and 1.73 BM
(4) 4.90, 0 and 2.83 BM
Ans. (3)
Sol. [FeCl] Fe²? 3d? → 4 unpaired electron. as Cl? in a weak field liquid.
μ_spin = √24 BM = 4.9 BM
[Co(C?O?)]³? Co³? 3d? → for Co³? with coordination no. 6 C?O?²? is strong field ligand & causes pairing & hence no. unpaired electron
μ_spin = 0
[MnO?]²? Mn?? it has one unpaired electron.
μ_spin = √3 BM

13. Which of the following reagent is suitable for the preparation of the product in the above reaction.
(1) Red P + Cl?
(2) NH?-NH?/C?H?ONa
(3) Ni/H?
(4) NaBH?

14. The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH ?
(1)
(2)
(3)
(4)
Ans. (3)
Sol.
Orange bright dye.

15. What is the correct sequence of reagents used for converting nitrobenzene into m-dibromobenzene ?
Ans. (4)

16. The correct shape and I-I-I bond angles respectively in I?? ion are :
(1) Trigonal planar; 120°
(2) Distorted trigonal planar; 135° and 90°
(3) Linear; 180°
(4) T-shaped; 180° and 90°
Ans. (3)
Sol. I?? sp³d hybridisation (2BP + 3L.P.) Linear geometry

17. What is the correct order of the following elements with respect to their density ?
(1) Cr < Fe < Co < Cu < Zn
(2) Cr < Zn < Co < Cu < Fe
(3) Zn < Cu < Co < Fe < Cr
(4) Zn < Cr < Fe < Co < Cu
Ans. (4)
Sol. Fact Based Density depend on many factor like atomic mass. atomic radius and packing efficiency.

18. Match List-I and List-II.
List-I List-II
(a) R-C-Cl → R-CHO (i) Br?/NaOH
(b) R-CH?-COOH → R-CH-COOH (ii) H?/Pd-BaSO?
Cl O
(c) R-C-NH? → R-NH? (iii) Zn(Hg)/Conc. HCl
O
(d) R-C-CH? → R-CH?-CH? (iv) Cl?/Red P, H?O
Choose the correct answer from the options given below :
(1) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Ans. (4)
O
(a) R-C-Cl H?/Pd-BaSO? → R-CHO (Rosemount reaction)
(b) R-CH?-COOH Cl/Red P, H?O → R-CH-COOH (HVZ reaction)
Cl O
(c) R-C-NH? Br?/NaOH → R-NH? (Hoffmann Bromamide reaction)
O
(d) R-C-CH? Zn(Hg)/conc.HCl → R-CH?-CH? (Clemensen reaction)

19. In polymer Buna-S ; 'S' stands for :
(1) Styrene
(2) Sulphur
(3) Strength
(4) Sulphonation
Ans. (1)
Sol. Buna-S is the co-polymer of buta-1, 3 diene & styrene.

20. Most suitable salt which can be used for efficient clotting of blood will be :
(1) Mg(HCO?)?
(2) FeSO?
(3) NaHCO?
(4) FeCl?

1. The magnitude of the change in oxidising power of the MnO??/Mn²? couple is × × 10?? V, if the H? concentration is decreased from 1M to 10?? M at 25°C. (Assume concentration of MnO?? and Mn²? to be same on change in H? concentration). The value of × is (Rounded off to the nearest integer)
Given: 2.303RT/F = 0.059
Ans. 3776
Sol. 5e? + MnO?? + 8H? → Mn²? + 4H?O
Q = [Mn²?]/([H?]?[MnO??]) ⇒ E? = E° - 0.059/5 log(Q?)
E? = E° - 0.059/5 log(Q?) ⇒ E? - E? = 0.059/5 log(Q?/Q?)
Q = 0.059/5 log([H?]?/[H?]??) ⇒ Q = 0.059/5 log(10??/1)?
(E? - E?) = 0.059/5 × (-32) ⇒ [E? - E?] = 32 × 0.059/5 = × × 10??
Q = (32 × 590)/5 × 10?? = × × 10?? ⇒ Q = 3776 × 10?? Q = 3776

2. Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is
(1) α-sulphur
(2) β-sulphur
(3) S?-form
Ans. (1)
Sol. S? is like O? ;e paramagnetic as per molecular orbital theory.

3. C?H? freezes at 5.5°C. The temperature at which a solution of 10 g of C?H?? in 200 g of C?H? freeze is °C. (The molal freezing point depression constant of C?H? is) 5.12°C/m
Sol. ΔT_f = i × K_f × m
= (1) × 5.12 × (10/58)/200 × 1000 ⇒ ΔT_f = (5.12 × 50)/58 = 4.414
T_f(solution) = T_K(solvent) - ΔT_f
= 5.5 - 4.414
= 1.086°C
≈ 1.09°C = 1 (nearest integer)

4. 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)
Ans. 5
Sol. T = 50°C = 323.15 K
P = 740 mm of Hg = 740/760 atm
V = ?
moles (n) = 4.75/26
V = 4.75/26 × (0.0821 × 323.15)/740 × 760
V = 4.97 = 5 lit

5. The solubility product of PbI? is 8.0 × 10??. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10?? mol/L. The value of x is (Rounded off to the nearest integer) [Given √2 = 1.41]
Ans. 141
Sol. K_sp[PbI?] = 8 × 10??
PbI?(s) ? Pb²?(aq) + 2I?(aq)
S + 0.1 2S
K_sp = [Pb²?][I?]²
8 × 10?? = (S + 0.1)(2S)² ⇒ 8 × 10?? = 0.1 × 4S²
⇒ S² = 2 × 10??
S = 1.414 × 10?? mol/Lit
⇒ x × 10?? mol/Lit ∴ x = 141.4 = 141

6. The total number of amines among the following which can be synthesized by Gabriel synthesis is
(1)
(2) CH?CH?NH?
(3)
(4)
Ans. (3)
Sol. Only aliphatic amines can be prepared by Gabriel synthesis.

7. 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.
Ans. 243
Sol. Ph-NH? Ph-NH-C-CH? (C?H?N) (Acetanilide)(C?H?NO) Molar mass = 93 Molar mass = 135
93 g Aniline produce 135 g acetanilide
1.86 g produce (135 × 1.86)/93 = 2.70g
At 10% loss, 90% product will be formed after purification.
Amount of product obtained = (2.70 × 90)/100 = 2.43g = 243 × 10?² g

8. The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O? for complete oxidation and produces 4 times its own volume of CO? is C_xH_y. The value of y is
Ans. 8
Sol. C_xH_y + 6O? → 4CO? + y/2 H?O
Applying POAC on 'O' atoms
6 × 2 = 4 × 2 + y/2 × 1
y/2 = 4 ⇒ y = 8

9. Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C. After 9h, the fraction of sucrose remaining is f. The value of log??(1/f) is × 10?² (Rounded off to the nearest integer) [Assume: ln 10 = 2.303, ln 2 = 0.693]
Ans. 81
Sol. Sucrose → Hydrolysis → Glucose + Fructose
t?/? = 3.33h = 10/3 h ⇒ C_t = C_o/2^(t/t?/?)
Fraction of sucrose remaining f = f = C_t/C_o = 1/2^(t/t?/?)
1/f = 2^(t/t?/?)
log(1/f) = log(2^(t/t?/?)) = t/t?/? log(2)
= 9/(10/3) × 0.3 = 8.1/10 = 0.81 = × × 10?² × × 81

10. Assuming ideal behaviour, the magnitude of log K for the following reaction at 25°C is × × 10?¹ The value of × is (Integer answer)
3HC≡CH(g) ? C?H?(g)
[Given: Δ_fG°(HC≡CH) = -2.04 × 10?] mol?¹; Δ_fG°(C?H?) = -1.24 × 10? J mol?¹ R = 8.314 J K?¹ mol?¹]
Ans. 855
Sol. 3HC≡CH(g) ? C?H?(?)
ΔG_r° = ΔG_r°[C?H?(?)] - 3 × ΔG_r°[HC≡CH]
= [-1.24 × 10? - 3 × (-2.04 × 10?)]
= 4.88 × 10? J/mol
ΔG_r° = -RT ln(K_eq)
log(K_eq) = -ΔG°/(2.303RT)
= -4.88 × 10?/(2.303 × 8.314 × 298)
= -8.55 × 10¹ = 855 × 10?¹

24th Feb. 2021 | Shift - 2
MATHEMATICS

1. 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 :
(1) 86
(2) 88
(3) 84
(4) 90
Ans. (2)
Sol. P(a,6,9), Q(20,b,-a-9)
mid point of PQ = ((a+20)/2, (b+6)/2, -a/2)
lie on line
((a+20)/2 - 3)/7 = ((b+6)/2 - 2)/5 = (-a/2 - 1)/(-9)
(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

2. 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:
(1) (9, 12)
(2) (6, 9)
(3) (3, 6)
(4) (0, 3)

4. The probability that two randomly selected subsets of the set {1,2,3,4,5} have exactly two elements in their intersection, is:
(1) 65/2?
(2) 135/2?
(3) 65/2?
(4) 35/2?
Ans. (2)
Sol. Required probability
= (?C? × 3³)/4?
= (10 × 27)/2¹? = 135/2?

5. The vector equation of the plane passing through the intersection of the planes r·(i+j+k)=1 and r·(i-2j)=-2, and the point (1,0,2) is :
(1) r·(i-7j+3k)=7/3
(2) r·(i+7j+3k)=7
(3) r·(3i+7j+3k)=7
(4) r·(i+7j+3k)=7/3
Ans. (2)
Sol. Plane passing through intersection of plane is
{r·(i+j+k)-1} + λ{r·(i-2j)+2} = 0
Passes through i+2k, we get
(3-1) + λ(1+2) = 0 ⇒ λ = -2/3
⇒ r·(i+7j+3k) = 7

6. If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x - 1, then the co-ordinates of P are :
(1) (-2,8)
(2) (1,5)
(3) (3,13)
(4) (2,8)
Ans. (4)
Sol. dy/dx|_p = 4
∴ 2x? = 4
⇒ x? = 2
∴ Point will be (2,8)

7. 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:
(1) 71/256
(2) 69/256
(3) 69/256
(4) 71/256
Ans. (4)
Sol. 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
c = 3/2
Quadratic Equation is 4x² + 11/4 x + 1 = 0
The value of (α+β)² - 3αβ = 121/256 - 3/4 = -71/256

8. The value of the integral, ∫?³ [x² - 2x - 2] dx, where [x] denotes the greatest integer less than or equal to x, is:
(1) -4
(2) -5
(3) -√2 - √3 - 1
(4) -√2 - √3 + 1
Ans. (3)
Sol. 1 = ∫?³ -3dx + ∫?³ [(x-1)²] dx
Put x - 1 = t; dx = dt
1 = (-6) + ∫?² [t²] dt
1 = -6 + ∫?¹ 0dt + ∫?^√2 1dt + ∫_√2^√3 2dt + ∫_√3² 3dt
1 = -6 + (√2 - 1) + 2√3 - 2√2 + 6 - 3√3
1 = -1 - √2 - √3

9. Let f: R → R be defined as
Let A = {x ∈ R : f is increasing}. Then A is equal to :
(1) (-5, -4) ∪ (4, ∞)
(2) (-5, ∞)
(3) (-∞, -5) ∪ (4, ∞)
(4) (-∞, -5) ∪ (-4, ∞)
Ans. (1)
Hence, f(x) is monotonically increasing in interval (-5, -4) ∪ (4, ∞)

10. 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 :
(1) a = 1, b = 1, c = 0
(3) a = 1, b = 0, c = 1
Ans. (1)
Sol. 2 = a + b + c ...(i)
dy/dx = 2ax + b ⇒ dy/dx|_(0,0) = 1
⇒ b = 1 ⇒ a + c = 1
(0,0) lie on curve
∴ c = 0, a = 1

11. The negation of the statement ~p ∧ (p ∨ q) is :
(1) ~p ∧ q
(2) p ∧ ~q
(3) ~p ∨ q
(4) p ∨ ~q
Ans. (4)
Sol.
∴ ~p ∧ (p ∨ q) ≡ p ∨ ~q

12. 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?
(1) (B) and (E) only
(2) (C) and (D) only
(3) (A) and (D) only
(4) (A) and (E) only
Ans. (3)
Sol. x - 2y + 0.z = 1
x - y + kz = -2
0.x + ky + 4z = 6
1 -2 0
1 -1 k
0 k 4
For unique solution 4 - k² ≠ 0
k ≠ ±2
For k = 2
x - 2y + 0.2 = 1
x - y + 2z = -2
0.x + 2y + 4z = 6
Δx = -48 ≠ 0
For k = 2 Δx ≠ 0
For k = 2 ; The system has no solution

13. For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2) ?
(1) x² + 9y² = 9
(2) 2x² - 18y² = 9
(3) y² = 1/(6√3) x
(4) x² + y² = 7
Ans. (1)
Sol. Tangent to x² + 9y² = 9 at point (3√3/2, 1/2) is x(3√3/2) + 9y(1/2) = 9
3√3x + 9y = 18 ⇒ x + √3y = 2√3
⇒ option (1) is true

14. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:
(1) 1200√3 m
(2) 1800√3 m
(3) 3600√3 m
(4) 2400√3 m
Ans. (1)
Sol. v = 432 × 1000/(60×60) m/sec = 120 m/sec
Distance AB = v × 20 = 2400 meter
In Δ PAC
tan 60° = h/PC ⇒ PC = h/√3
In Δ PBD
tan 30° = h/PD ⇒ PD = √3h
PD = PC + CD
√3h = h/√3 + 2400 ⇒ 2h/√3 = 2400
h = 1200√3 meter

15. For the statements p and q, consider the following compound statements:
(a)(~q ∧ (p → q)) → ~p
(b)((p ∨ q)) ∧ ~p) → p
Then which of the following statements is correct?
(1) (a) is a tautology but not (b)
(2) (a) and (b) both are not tautologies.
(3) (a) and (b) both are tautologies.
(4) (b) is a tautology but not (a).
Ans. (3)
(a) is tautologies
(b) is tautologies
a & b are both tautologies.

16. 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 :
(1) a unique solution
(2) exactly two solutions
(3) infinitely many solutions
(4) no solution

18. If a curve y = f(x) passes through the point (1,2) and satisfies x dy/dx + y = bx?, then for what value of b, ∫?² f(x)dx = 62/5 ?
(1) 5
(2) 62/5
(3) 31/5
(4) 10
Ans. (4)
Sol. dy/dx + y/x = bx³, I.F. = e^{∫dx/x} = x
∴ yx = ∫bx?dx = bx?/5 + C
Passes through (1,2), we get
2 = b/5 + C ...(1)
Also, ∫?² (bx?/5 + C/x) dx = 62/5
⇒ b/25 × 32 + C ln2 - b/25 = 62/5 ⇒ C = 0 & b = 10

19. The area of the region : R = {(x,y) : 5x² ≤ y ≤ 2x² + 9} is:
(1) 9√3 square units
(2) 12√3 square units
(3) 11√3 square units
(4) 6√3 square units
Ans. (2)

20. 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 ∫?² f(x)dx is:
(1) 1 + e²
(2) 1 - e²
(3) 2(1 - e²)
(4) 2(1 + e²)
Ans. (1)
Sol. f'(x) = f'(2-x)
On integrating both side 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)
I = ∫?² f(x)dx = ∫?¹ {f(x) + f(2-x)} dx = (1 + e²)

Section B

1. The number of the real roots of the equation (x+1)² + |x-5| = 27/4 is
Ans. 2
Sol. x ≥ 5
(x+1)² + (x-5) = 27/4
⇒ x² + 3x - 4 = 27/4
⇒ x² + 3x - 43/4 = 0
⇒ 4x² + 12x - 43 = 0

3. If a + α = 1, b + β = 2 and af(x) + αf(1/x) = bx + β/x, x ≠ 0, then the value of the expression (f(x) + f(1/x))/(x + 1/x) is
Ans. 2
Sol.
af(x) + αf(1/x) = bx + β/x
x → 1/x
af(1/x) + αf(x) = b/x + βx
(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

4. If the variance of 10 natural numbers 1,1,1,...,1,k is less than 10, then the maximum possible value of k is
Ans. 11
Sol. σ² = Σx²/n - (Σx/n)²
σ² = (9+k²)/10 - ((9+k)/10)² < 10
(90 + k²)10 - (81 + k² + 8k) < 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 integral value of k = 11

5. 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
Ans. 1
Sol. (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)
Distance between skew lines = |(a?-a?) b? b?| / |b? × b?|
| λ 1/2+2λ -λ |
| 2 1 -1 |
| 1 1 1 |
| i j k |
| 2 1 -1 |
| 1 1 1 |
= |-5λ - 3/2|/√14 = √7/(2√2) (given)
|10λ + 3| = 7 ⇒ λ = -1
⇒ |λ| = 1

6. Let i = √-1. If ((-1+i√3)²¹)/((1-i)²?) + ((1+i√3)²¹)/((1+i)²?) = k, and n = [|k|] be the greatest integral part of |k|. Then ∑_{j=0}^{n+5} (j+5)² - ∑_{j=0}^{n+5} (j+5) is equal to______.
Ans. 310
Sol.
(2e^{i2π/3}/(√2 e^{-iπ/4}))²¹ / ( (√2 e^{iπ/4})²? ) + ...
⇒ 2²¹e^{i14π}/(2¹² e^{-i6π}) + 2²¹(e^{i7π})/(2¹²(e^{i6π}))
⇒ 2? e^{i(20π)} + 2? e^{iπ}
⇒ 2? + 2?(-1) = 0
n = 0
∑_{j=0}^5 (j+5)² - ∑_{j=0}^5 (j+5)

7. 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
Ans. 56.25
Sol. Let P(h,k)
Given
PA = 3PB
PA² = 9PB²
⇒ (h-5)² + k² = 9[(h+5)² + k²]
⇒ 8h² + 8k² + 100h + 200 = 0
Locus
x² + y² + (25/2)x + 25 = 0
∴ c = (-25/4, 0)
∴ r² = (-25/4)² - 25
= 625/16 - 25
= 225/16
∴ 4r² = 4 × 225/16 = 225/4 = 56.25

8. For integers n and r, let (n r) = { nCr, if n ≥ r ≥ 0; 0, otherwise }
The maximum value of k for which the sum ∑_{i=0}^k (10 i)(15 k-i) + ∑_{i=0}^{k+1} (12 i)(13 k+1-i) exists, is equal to
Ans. 12
Sol. (1+x)¹? = ¹?C? + ¹?C?x + ... + ¹?C??x¹?
(1+x)¹? = ¹?C? + ¹?C?x + ... + ¹?C_{k-1}x^{k-1} + ¹?C_k x^k + ¹?C_{k+1}x^{k+1} + ... + ¹?C??x¹?
∑_{i=0}^k (10Ci)(15C_{k-i}) = ¹?C?.¹?C_k + ¹?C?.¹?C_{k-1} + ... + ¹?C_k.¹?C?
Coefficient of x^k in (1+x)²? = ²?C_k
∑_{i=0}^{k+1} (12Ci)(13C_{k+1-i}) = ¹²C?.¹³C_{k+1} + ... + ¹²C_{k+1}.¹³C?
Coefficient of x^{k+1} in (1+x)²? = ²?C_{k+1}
²?C_k + ²?C_{k+1} = ²?C_{k+1}
For maximum value
k + 1 = 13
K = 12

9. 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 ______.
Ans. 3
Sol. a, ar, ar², ar³
a + ar + ar² + ar³ = 65/12
1/a + 1/ar + 1/ar² + 1/ar³ = 65/18

10. 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
Ans. 1225
Sol. Equation of normal at P
(y-7) = ((7-3)/(5-2))(x-5)
3y - 21 = 4x - 20
⇒ 4x - 3y + 1 = 0
⇒ M(-1/4, 0)
Equation of tangent at P
4y - 28 = -3x + 15 ⇒ 3x + 4y = 43 (ii) ⇒ N(43/3, 0)
Hence ar(ΔPMN) = 1/2 × MN × 7
λ = 1/2 × 175/12 × 7
⇒ 24λ = 1225

JEE MAINS 2021 Solved Question paper Free PDF Download Link All sets

2012 to 2026 JEE mains last 15 Year solved question paper Free PDF download link

JEE MAINS 2021 AUGUST Solved Question paper

JEE MAINS 2021 MORNING SESSION SOLVED QUESTION PAPER