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JEE Main 7 January 2020 Shift 1 Question Paper with Answers

7 January 2020 · January session · 74 questions

74 of the 75 questions from the JEE Main 7 January 2020 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
25
Chemistry
25
Mathematics
24

Physics — JEE Main 7 January 2020 Shift 1

Q1·PhysicsSingle correct
A polarizer-analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer-analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero is
  1. (A)18.4°18.4°18.4°
  2. (B)45°45°45°
  3. (C)71.6°71.6°71.6°
  4. (D)90°90°90°

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
Two infinite planes each with uniform surface charge density +σ+\sigma+σ are kept in such a way that the angle between them is 30°. The electric field in the region shown between them is given by:
  1. (A)σ2ε0[(1−32)y^−x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1-\dfrac{\sqrt{3}}{2}\right)\hat{y}-\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1−23​​)y^​−2x^​]
  2. (B)σ2ε0[(1+3)y^−x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1+\sqrt{3}\right)\hat{y}-\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1+3​)y^​−2x^​]
  3. (C)σ2ε0[(1+3)y^+x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1+\sqrt{3}\right)\hat{y}+\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1+3​)y^​+2x^​]
  4. (D)σε0[(1+32)y^+x^2]\dfrac{\sigma}{\varepsilon_0}\left[\left(1+\dfrac{\sqrt{3}}{2}\right)\hat{y}+\dfrac{\hat{x}}{2}\right]ε0​σ​[(1+23​​)y^​+2x^​]

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A long solenoid of radius R carries a time (t) dependent current I (t) = I0t(1−t)I_0 t (1-t)I0​t(1−t). A ring of radius 2R is placed cordially near its middle. During the time internal 0≤t≤10 \le t \le 10≤t≤1, the induced current (IRI_RIR​) and the induced EMF (VRV_RVR​) in the ring changes as:
  1. (A)At t = 0.25 direction of IRI_RIR​ reverses and VRV_RVR​ is maximum.
  2. (B)Direction of IRI_RIR​ remains unchanged and VRV_RVR​ is zero at t = 0.25
  3. (C)Direction of IRI_RIR​ remains unchanged and VRV_RVR​ is maximum at t = 0.5
  4. (D)At t = 0.5 direction of IRI_RIR​ reverses and VRV_RVR​ is zero.

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
Visible light of wavelength 6000×10−86000 \times 10^{-8}6000×10−8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at θ1\theta_1θ1​ then θ1\theta_1θ1​ is close to:
  1. (A)20°20°20°
  2. (B)30°30°30°
  3. (C)25°25°25°
  4. (D)45°45°45°

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(n) = k(1 + \alpha n) where 'x' is he distance measured from one of the plates. If (α\alphaαd) << 1, the total capacitance of the system is best given by the expression:
  1. (A)AKε0d(1+αd2)\dfrac{AK\varepsilon_0}{d}\left(1+\dfrac{\alpha d}{2}\right)dAKε0​​(1+2αd​)
  2. (B)Aε0Kd(1+α2d22)\dfrac{A\varepsilon_0 K}{d}\left(1+\dfrac{\alpha^2 d^2}{2}\right)dAε0​K​(1+2α2d2​)
  3. (C)AKε0d(1+αd)\dfrac{AK\varepsilon_0}{d}(1+\alpha d)dAKε0​​(1+αd)
  4. (D)Aε0Kd[1+(αd2)2]\dfrac{A\varepsilon_0 K}{d}\left[1+\left(\dfrac{\alpha d}{2}\right)^2\right]dAε0​K​[1+(2αd​)2]

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
The current I1I_1I1​ (in A) flowing through 1 Ω\OmegaΩ resistor in the following circuit is
  1. (A)0.25
  2. (B)0.4
  3. (C)0.2
  4. (D)0.5

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
The radius of gyration of a uniform rod of length ℓ\ellℓ, about an axis passing through a point ℓ4\dfrac{\ell}{4}4ℓ​ away from the centre of the rod, an perpendicular to it is:
  1. (A)18ℓ\dfrac{1}{8}\ell81​ℓ
  2. (B)748ℓ\sqrt{\dfrac{7}{48}}\ell487​​ℓ
  3. (C)14ℓ\dfrac{1}{4}\ell41​ℓ
  4. (D)38ℓ\sqrt{\dfrac{3}{8}}\ell83​​ℓ

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
Three point particles of masses 10 kg, 1.5 kg and 2.5 kg are placed at there corners of a right angle triangle of sides 4.0 cm, 3.0 ;cm and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:
  1. (A)0.6 cm right and 2.0 cm above 1 kg mass.
  2. (B)2.0 cm right and 0.9 cm above 1 kg mass.
  3. (C)1.5 cm right and 1.2 cm above 1 kg mass.
  4. (D)0.9 cm right and 2.0 cm above 1 kg mass.

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by ϕ\phiϕ. The magnetic flux through the area is given by ϕ0\phi_0ϕ0​. Which of the following is correct?
  1. (A)ϕi=−ϕ0\phi_i=-\phi_0ϕi​=−ϕ0​
  2. (B)ϕi>ϕ0\phi_i>\phi_0ϕi​>ϕ0​
  3. (C)ϕi=ϕ0\phi_i=\phi_0ϕi​=ϕ0​
  4. (D)ϕi<ϕ0\phi_i<\phi_0ϕi​<ϕ0​

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be
  1. (A)r34ghr\sqrt{\dfrac{3}{4gh}}r4gh3​​
  2. (B)1r2gh3\dfrac{1}{r}\sqrt{\dfrac{2gh}{3}}r1​32gh​​
  3. (C)r32ghr\sqrt{\dfrac{3}{2gh}}r2gh3​​
  4. (D)1r4gh3\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}r1​34gh​​

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye piece should be close to
  1. (A)22 mm
  2. (B)33 m
  3. (C)12 m
  4. (D)2 mm

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
Two moles of an ideal gas with CPCV=53\dfrac{C_P}{C_V}=\dfrac{5}{3}CV​CP​​=35​ are mixed with 3 moles of another ideal gas with CPCV=43\dfrac{C_P}{C_V}=\dfrac{4}{3}CV​CP​​=34​. The value of CPCV\dfrac{C_P}{C_V}CV​CP​​ for the mixture is:
  1. (A)1.42
  2. (B)1.47
  3. (C)1.45
  4. (D)1.50

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2mm^{2}mm2) is 90 ms−1ms^{-1}ms−1. If the young's modulus of wire is 16×101116 \times 10^{11}16×1011 Nm−2Nm^{-2}Nm−2, the extension of wire over its natural length is:
  1. (A)0.03 mm
  2. (B)0.01 mm
  3. (C)0.02 mm
  4. (D)0.04 mm

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
A litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ = 1.40, the work done by air is: (31.43^{1.4}31.4 = 4.6555) [Take air to be an ideal gas]
  1. (A)48 J
  2. (B)90.5 J
  3. (C)100.8 J
  4. (D)60.7 J

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass m10\frac{m}{10}10m​ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant ; M is the mass of the earth)
  1. (A)m20(u−2GM3R)2\frac{m}{20}\left(u-\sqrt{\frac{2GM}{3R}}\right)^{2}20m​(u−3R2GM​​)2
  2. (B)3m8(u+5GM6R)2\frac{3m}{8}\left(u+\sqrt{\frac{5GM}{6R}}\right)^{2}83m​(u+6R5GM​​)2
  3. (C)m20(u2−113200GMR)\frac{m}{20}\left(u^{2}-\frac{113}{200}\frac{GM}{R}\right)20m​(u2−200113​RGM​)
  4. (D)5m(u2−119200GMR)5m\left(u^{2}-\frac{119}{200}\frac{GM}{R}\right)5m(u2−200119​RGM​)

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6×10−161.6 \times 10^{-16}1.6×10−16 s. The frequency of revolution of the electron is its first excited state (in s−1^{-1}−1) is
  1. (A)7.8×10147.8 \times 10^{14}7.8×1014
  2. (B)1.6×10141.6 \times 10^{14}1.6×1014
  3. (C)6.2×10156.2 \times 10^{15}6.2×1015
  4. (D)5.6×10125.6 \times 10^{12}5.6×1012

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the connect equivalence would be:
  1. (A)L ↔\leftrightarrow↔ 1b\frac{1}{b}b1​, C ↔\leftrightarrow↔ 1m\frac{1}{m}m1​, R ↔\leftrightarrow↔ 1k\frac{1}{k}k1​
  2. (B)L ↔\leftrightarrow↔ k, C ↔\leftrightarrow↔ b, R ↔\leftrightarrow↔ m
  3. (C)L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ 1k\frac{1}{k}k1​, R ↔\leftrightarrow↔ b
  4. (D)L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ k, R ↔\leftrightarrow↔ b

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
If the magnetic field in a plane electromagnetic wav is given by B⃗=3×10−8sin⁡(1.6×103x+48×1010t)j^\vec{B} = 3 \times 10^{-8} \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t)\hat{j}B=3×10−8sin(1.6×103x+48×1010t)j^​ T, then what will be expression for electric field?
  1. (A)E⃗=(60sin⁡(1.6×103x+48×1010t)k^ V/m)\vec{E} = \left(60 \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t)\hat{k} \text{ V/m}\right)E=(60sin(1.6×103x+48×1010t)k^ V/m)
  2. (B)E⃗=(3×10−8sin⁡(1.6×103x+48×1010t)i^ V/m)\vec{E} = \left(3 \times 10^{-8} \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t)\hat{i} \text{ V/m}\right)E=(3×10−8sin(1.6×103x+48×1010t)i^ V/m)
  3. (C)E⃗=(9sin⁡(1.6×103x+48×1010t)k^ V/m)\vec{E} = \left(9 \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t)\hat{k} \text{ V/m}\right)E=(9sin(1.6×103x+48×1010t)k^ V/m)
  4. (D)E⃗=(3×10−8sin⁡(1.6×103x+48×1010t)j^ V/m)\vec{E} = \left(3 \times 10^{-8} \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t)\hat{j} \text{ V/m}\right)E=(3×10−8sin(1.6×103x+48×1010t)j^​ V/m)

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
Which of the following given a reversible operation?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms−2^{-2}−2)
  1. (A)1.5 ms−1^{-1}−1
  2. (B)2.0 ms−1^{-1}−1
  3. (C)1.7 m/s−1^{-1}−1
  4. (D)1.9 m/s−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q21·PhysicsNumerical
A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is _________.

Correct answer: 600

Step-by-step solution →
Q22·PhysicsNumerical
A loop ABCDEFA of straight edges has six corner points A (0, 0, 0), B (5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region in B⃗=(3i^+4k^)\vec{B} = (3\hat{i} + 4\hat{k})B=(3i^+4k^)T. The quantity of flux through the loop ABCDEFA (in Wb) is _________.

Correct answer: 175

Step-by-step solution →
Q23·PhysicsNumerical
A non-isotropic solid metal cube has coefficients of linear expansion as: 5×10−55 \times 10^{-5}5×10−5/°C along the x = axis and 5×10−65 \times 10^{-6}5×10−6/°C along the y and the z-axis. If coefficient of volume expansion of the solid C ×10−6\times 10^{-6}×10−6/°C then the value of C is _________.

Correct answer: 60

Step-by-step solution →
Q24·PhysicsNumerical
A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest of a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it leading its highest point P(height 1 m) the kinetic energy of the particle (in J) is: (Figure drawn is schematic and not to scale (take g = 10 ms−2^{-2}−2)

Correct answer: 10

Step-by-step solution →
Q25·PhysicsNumerical
A beam of electromagnetic radiation of intensity 6.4×10−56.4 \times 10^{-5}6.4×10−5 W/cm2^{2}2 is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function φ = 2 eV) of surface area 1 cm2^{2}2. If one in 10310^{3}103 photons ejects an electron, total number of electrons ejected in is 10x10^{x}10x. (hch_chc​ = 1240 eVnm, 1 eV = 1.6×10−191.6 \times 10^{-19}1.6×10−19 J), then x is _________.

Correct answer: 11

Step-by-step solution →

Chemistry — JEE Main 7 January 2020 Shift 1

Q26·ChemistrySingle correct
The relative strength of interionic/intermolecular forces in decreasing order is
  1. (A)ion-dipole > dipole-dipole > ion-ion
  2. (B)dipole-dipole > ion-dipole > ion- ion
  3. (C)ion-ion > ion-dipole > dipole-dipole
  4. (D)ion-dipole > ion-ion > dipole-dipole

Correct answer: (C)

Step-by-step solution →
Q27·ChemistrySingle correct
Oxidation number of potassium in K2_22​O, K2_22​O2_22​ and KO2_22​ respectively is
  1. (A)+1, +2 and +4
  2. (B)+2, +1 and +½
  3. (C)+1, +1 and +1
  4. (D)+1, +4 and +2

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correct
At 35°C, the vapour pressure of CS2_22​ is 512 mm of Hg and that of acetone is 344 mm of Hg. A solution of CS2_22​ in acetone has a total vapour pressure of 600 mm of Hg. The false statement among the following is
  1. (A)CS2_22​ and acetone are less attracted to each other than to themselves
  2. (B)Heat must be absorbed in order to produce the solution at 35°C
  3. (C)Raoult's law is not obeyed by this system
  4. (D)a mixture of 100 mL CS2_22​ and 100 mL acetone has a volume of < 200 mL

Correct answer: (D)

Step-by-step solution →
Q29·ChemistrySingle correct
The atomic radius of Ag is closest to
  1. (A)Ni
  2. (B)Cu
  3. (C)Au
  4. (D)Hg

Correct answer: (C)

Step-by-step solution →
Q30·ChemistrySingle correct
The dipole moments of CCl4_44​, CHCl3_33​ ad CH4_44​ are in the order
  1. (A)CHCl3_33​ < CH4_44​ = CCl4_44​
  2. (B)CCl4_44​ < CH4_44​ < CHCl3_33​
  3. (C)CH4_44​ = CCl4_44​ < CHCl3_33​
  4. (D)CH4_44​ < CCl4_44​ < CHCl3_33​

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is
  1. (A)less efficient as it exchanges only anions
  2. (B)more efficient as it can exchange only cations
  3. (C)less efficient as the resin cannot be regenerated
  4. (D)more efficient as it can exchange both cation as well as anions

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
Amongst the following statements, that which was not proposed by Dalton was
  1. (A)matter consists of invisible atoms
  2. (B)when gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same temperature & pressure
  3. (C)chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction
  4. (D)All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
The increasing order of pKb_bb​ for the following compounds will be (a) NH2_22​-CH=NH2_22​ (b) [structure shown] (c) CH3_33​NHCH3_33​
  1. (A)b < c < a
  2. (B)c < a < b
  3. (C)b < a < c
  4. (D)a < b < c

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
What is the product of following reaction? Hex − 3 − ynal →(ii) PBr3 (iii) Mg/ether (iv) CO2/H3O+(i) NaBH4\xrightarrow[\text{(ii) PBr}_3\ \text{(iii) Mg/ether}\ \text{(iv) CO}_2\text{/H}_3\text{O}^+]{\text{(i) NaBH}_4}(i) NaBH4​(ii) PBr3​ (iii) Mg/ether (iv) CO2​/H3​O+​ ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The number of orbitals associated with quantum number n = 5, ms_ss​ = +12\frac{1}{2}21​ is
  1. (A)15
  2. (B)11
  3. (C)50
  4. (D)25

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
The purest form of commercial iron is
  1. (A)scrap iron and pig iron
  2. (B)wrought iron
  3. (C)cast iron
  4. (D)pig iron

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
The theory that can completely/properly explain the nature of bonding in [Ni(CO)4_44​] is
  1. (A)Werner's theory
  2. (B)Crystal field theory
  3. (C)Molecular orbital theory
  4. (D)Valence bond theory

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
The IUPAC name of the complex is [Pt(NH3_33​)2_22​Cl(NH2_22​CH3_33​)]Cl is
  1. (A)Diamminechlorido(aminomethane)platinum(II)chloride
  2. (B)Diamminechlorido(methanamine)platinum(II)chloride
  3. (C)Diammine(methanamine)chlorido platinum(II)chloride
  4. (D)Bisammine(methanamine)chloride platinum(II)chloride

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
1-methyl ethylene oxide when treated with an excess of HBr, produces
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Consider the following reaction: →OH−\xrightarrow{OH^{-}}OH−​ 'X' The product 'X' is used
  1. (A)in protein estimation as an alternative to ninhydrin
  2. (B)as food grade colourant
  3. (C)in laboratory test for phenols
  4. (D)in acid-base titration as an indicator

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Match the following
Column – IColumn – II
i.Riboflamina.Beriberi
ii.Thiamineb.Scurvy
iii.Pyridoxinec.Cheilosis
iv.Ascorbic acidd.Convulsions
  1. (A)i → c, ii → a, iii → d, iv → b
  2. (B)i → c, ii → d, iii → a, iv → b
  3. (C)i → a, ii → d, iii → c, iv → b
  4. (D)i → d, ii → b, iii → a, iv → c

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
Given that the standard potential (E°) of Cu2+^{2+}2+/Cu and Cu+^{+}+/Cu are 0.34 V and 0.522 V respectively, the E° of Cu2+^{2+}2+/Cu+^{+}+ is
  1. (A)-0.158 V
  2. (B)0.182 V
  3. (C)+0.158 V
  4. (D)-0.182 V

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3_33​ to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fraction A, B and C contain respectively
  1. (A)m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline
  2. (B)m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline
  3. (C)m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol
  4. (D)m-chloroaniline, m-chlorobenzoic acid and m-chlorophenol

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
The order of electron gain enthalpy in kJ/mol of fluorine, chlorine, bromine and iodine, respectively are
  1. (A)-333, -325, -349 and -296
  2. (B)-333, -349, -325 and -296
  3. (C)-349, -333, -325 and -296
  4. (D)-296, -325, -333 and -349

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Consider the following reactions, which of these reaction(s) will not produce Saytzeff product? (a) (CH3_33​)3_33​CCH(OH)CH3_33​ →cons. H2SO4\xrightarrow{cons.\ H_2SO_4}cons. H2​SO4​​ (b) (CH3_33​)2_22​CHCH(Br)CH3_33​ →cons. KOH\xrightarrow{cons.\ KOH}cons. KOH​ (c) (CH3_33​)2_22​CHCH(Br)CH3_33​ →(CH3)3O⊖K⊕\xrightarrow{(CH_3)_3O^{\ominus}K^{\oplus}}(CH3​)3​O⊖K⊕​ (d) (CH3_33​)2_22​C(OH)CH2_22​CHO →Δ\xrightarrow{\Delta}Δ​ Which of these reaction(s) will not produce Saytzeff product?
  1. (A)(b) and (d)
  2. (B)(c) only
  3. (C)(a), (c) and (d)
  4. (D)(d) only

Correct answer: (B)

Step-by-step solution →
Q46·ChemistryNumerical
Two solutions A and B, each of 100 L was made by dissolution of 4g of NaOH and 9.8 g of H2_22​SO4_44​ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

Correct answer: 10.60

Step-by-step solution →
Q47·ChemistryNumerical
During the nuclear explosion, one of the product of 90^{90}90Sr with half life of 6.93 years. If 1 μ\muμg of 90^{90}90Sr absorbed in the bones of newly born baby in place of Ca, how much time, in years is required to reduce it by 90% if it is not lost metabolically

Correct answer: 23.03

Step-by-step solution →
Q48·ChemistryNumerical
Chlorine reacts with hot and concentrated NaOH and produces compound(X) and (Y). Compound(X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is

Correct answer: 1.67

Step-by-step solution →
Q49·ChemistryNumerical
The number of chiral carbons in chloramphenicol is

Correct answer: 2

Step-by-step solution →
Q50·ChemistryNumerical
For the reaction A(l) ⟶\longrightarrow⟶ 2B(g) ΔU\Delta UΔU = 2.1 Kcal, ΔS\Delta SΔS = 20 cal K−1^{-1}−1 at 300 K Hence ΔG\Delta GΔG in Kcal is

Correct answer: -2.70

Step-by-step solution →

Mathematics — JEE Main 7 January 2020 Shift 1

Q51·MathematicsSingle correct
Let the function, f:[−7,0]→Rf:[-7,0]\to Rf:[−7,0]→R be continuous on [−7,0][-7,0][−7,0] and differentiable on (−7,0)(-7,0)(−7,0). If f(−7)=−3f(-7)=-3f(−7)=−3 and f′(x)≤2f'(x)\le 2f′(x)≤2, for all x∈(−7,0)x\in(-7,0)x∈(−7,0), then for all such functions f, f(−1)+f(0)f(-1)+f(0)f(−1)+f(0) lies in the interval:
  1. (A)[−3,11][-3,11][−3,11]
  2. (B)(−∞,20](-\infty, 20](−∞,20]
  3. (C)[−6,20][-6,20][−6,20]
  4. (D)(−∞,11](-\infty,11](−∞,11]

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:
  1. (A)3\sqrt{3}3​
  2. (B)323\sqrt{2}32​
  3. (C)32\frac{3}{\sqrt{2}}2​3​
  4. (D)232\sqrt{3}23​

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
An unbiased coin is tossed 5 time. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value −1. Then the expected value of X, is
  1. (A)−316-\frac{3}{16}−163​
  2. (B)−18-\frac{1}{8}−81​
  3. (C)18\frac{1}{8}81​
  4. (D)316\frac{3}{16}163​

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is −12-\frac{1}{2}−21​, then the greatest number amongst them is:
  1. (A)7
  2. (B)212\frac{21}{2}221​
  3. (C)16
  4. (D)27

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correct
If the system of linear equations 2x+2ay+az=02x+2ay+az=02x+2ay+az=0, 2x+3by+bz=0,2x+4cy+cz=02x+3by+bz=0,2x+4cy+cz=02x+3by+bz=0,2x+4cy+cz=0 where a,b,c ∈R\in R∈R are non-zero and distinct; has a non-zero solution, then:
  1. (A)a+b+c=0a+b+c=0a+b+c=0
  2. (B)a, b, c are in A.P.
  3. (C)1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c}a1​,b1​,c1​ are in A.P.
  4. (D)a, b, c are in G.P.

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correct
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is:
  1. (A)(6, 5, 2)
  2. (B)(4, 3, 2)
  3. (C)(6, 5, −2)
  4. (D)(3, 4, −2)

Correct answer: (C)

Step-by-step solution →
Q57·MathematicsSingle correct
If Re(z−12z+i)=1Re\left(\frac{z-1}{2z+i}\right)=1Re(2z+iz−1​)=1, where z=x+iyz=x+iyz=x+iy, then the point (x, y) lies on a:
  1. (A)straight line whose slope is 32\frac{3}{2}23​
  2. (B)circle whose diameter is 52\frac{\sqrt{5}}{2}25​​
  3. (C)straight line whose slope is −23-\frac{2}{3}−32​
  4. (D)circle whose centre is at (−12,−32)\left(-\frac{1}{2},-\frac{3}{2}\right)(−21​,−23​)

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
If y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2αy(\alpha)=\sqrt{2\left(\frac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha}\right)+\frac{1}{\sin^{2}\alpha}}y(α)=2(1+tan2αtanα+cotα​)+sin2α1​​, α∈(3π4,π)\alpha\in\left(\frac{3\pi}{4},\pi\right)α∈(43π​,π) then dydα\frac{dy}{d\alpha}dαdy​ at α=5π6\alpha=\frac{5\pi}{6}α=65π​ is:
  1. (A)4
  2. (B)−14-\frac{1}{4}−41​
  3. (C)−4
  4. (D)43\frac{4}{3}34​

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
Let α\alphaα be a root of the equation x2+x+1=0x^{2}+x+1=0x2+x+1=0 and the matrix A=13[1111αα21α2α4]A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4} \end{bmatrix}A=3​1​​111​1αα2​1α2α4​​, then the matrix A31A^{31}A31 is equal to:
  1. (A)A3A^{3}A3
  2. (B)A
  3. (C)I3I_{3}I3​
  4. (D)A2A^{2}A2

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
If g(x)=x2+x−1g(x)=x^{2}+x-1g(x)=x2+x−1 and (gof)(x)=4x2−10x+5(gof)(x)=4x^{2}-10x+5(gof)(x)=4x2−10x+5, then f(54)f\left(\frac{5}{4}\right)f(45​) is equal to
  1. (A)32\frac{3}{2}23​
  2. (B)12\frac{1}{2}21​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
If y=mx+4y=mx+4y=mx+4 is a tangent to both the parabolas, y2=4xy^{2}=4xy2=4x and x2=2byx^{2}=2byx2=2by, then b is equal to:
  1. (A)−32
  2. (B)−128
  3. (C)−64
  4. (D)128

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
The logical statement (p⇒q)∧(q⇒∼p)(p\Rightarrow q)\wedge(q\Rightarrow\sim p)(p⇒q)∧(q⇒∼p) is equivalent to:
  1. (A)∼q\sim q∼q
  2. (B)p
  3. (C)q
  4. (D)∼p\sim p∼p

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
Let xk+yk=ak,(a,k>0)x^{k} + y^{k} = a^{k}, (a,k > 0)xk+yk=ak,(a,k>0) and dydx+(yx)13=0\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{3}} = 0dxdy​+(xy​)31​=0, then k is:
  1. (A)32\frac{3}{2}23​
  2. (B)43\frac{4}{3}34​
  3. (C)13\frac{1}{3}31​
  4. (D)23\frac{2}{3}32​

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
Let α\alphaα and β\betaβ be two real roots of the equation (k+1)tan⁡2x−2.λtan⁡x=(1−k)(k+1)\tan^{2}x - \sqrt{2}.\lambda \tan x = (1-k)(k+1)tan2x−2​.λtanx=(1−k), where k(≠−1)k(\ne -1)k(=−1) and λ\lambdaλ are real numbers. If tan⁡2(α+β)=50\tan^{2}(\alpha+\beta) = 50tan2(α+β)=50, then a value of λ\lambdaλ is:
  1. (A)10
  2. (B)5
  3. (C)525\sqrt{2}52​
  4. (D)10210\sqrt{2}102​

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation, ey(dydx−1)=exe^{y}\left(\frac{dy}{dx} - 1\right) = e^{x}ey(dxdy​−1)=ex such that y(0)=0y(0) = 0y(0)=0, then y(1)y(1)y(1) is equal to:
  1. (A)1+log⁡e21 + \log_{e} 21+loge​2
  2. (B)2e2e2e
  3. (C)log⁡e2\log_{e} 2loge​2
  4. (D)2+log⁡e22 + \log_{e} 22+loge​2

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
The area of the region, enclosed by the circle x2+y2=2x^{2} + y^{2} = 2x2+y2=2, which is not common to the region bounded by the parabola y2=xy^{2} = xy2=x and the straight line y=xy = xy=x, is :
  1. (A)13(12π−1)\frac{1}{3}(12\pi - 1)31​(12π−1)
  2. (B)13(6π−1)\frac{1}{3}(6\pi - 1)31​(6π−1)
  3. (C)16(12π−1)\frac{1}{6}(12\pi - 1)61​(12π−1)
  4. (D)16(24π−1)\frac{1}{6}(24\pi - 1)61​(24π−1)

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
A vector a⃗=αi^+2j^+βk^ (α,β∈R)\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k} \ (\alpha,\beta \in R)a=αi^+2j^​+βk^ (α,β∈R) lies in the plane of the vectors, b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}b=i^+j^​ and c⃗=i^−j^+4k^\vec{c} = \hat{i} - \hat{j} + 4\hat{k}c=i^−j^​+4k^. If a⃗\vec{a}a bisects the angle between b⃗\vec{b}b and c⃗\vec{c}c, then:
  1. (A)a⃗.k^+4=0\vec{a}.\hat{k} + 4 = 0a.k^+4=0
  2. (B)a⃗.k^+2=0\vec{a}.\hat{k} + 2 = 0a.k^+2=0
  3. (C)a⃗.i^+3=0\vec{a}.\hat{i} + 3 = 0a.i^+3=0
  4. (D)a⃗.i^+1=0\vec{a}.\hat{i} + 1 = 0a.i^+1=0

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
The greatest positive integer k, for which 49k+149^{k}+149k+1 is a factor of the sum 49125+49124+……+492+49+149^{125} + 49^{124} + \ldots\ldots + 49^{2} + 49 + 149125+49124+……+492+49+1, is:
  1. (A)65
  2. (B)63
  3. (C)32
  4. (D)60

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
Total number of 6 - digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is
  1. (A)565^{6}56
  2. (B)6!6!6!
  3. (C)52(6!)\frac{5}{2}(6!)25​(6!)
  4. (D)12(6!)\frac{1}{2}(6!)21​(6!)

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsNumerical
Let S be the set of points where the function, f(x)=∣2−∣x−3∣∣,x∈Rf(x) = \left|2 - |x-3|\right|, x \in Rf(x)=∣2−∣x−3∣∣,x∈R, is not differentiable. Then ∑x∈Sf(f(x))\sum\limits_{x \in S} f(f(x))x∈S∑​f(f(x)) is equal to _________.

Correct answer: 3

Step-by-step solution →
Q71·MathematicsNumerical
If the sum of the coefficients of all even powers of x in the product (1+x+x2+......+x2n)(1−x+x2−x3+.....+x2n)(1+x+x^{2}+......+x^{2n})(1-x+x^{2}-x^{3}+.....+x^{2n})(1+x+x2+......+x2n)(1−x+x2−x3+.....+x2n) is 61, then n is equal to

Correct answer: 30

Step-by-step solution →
Q72·MathematicsNumerical
If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to _________

Correct answer: 18

Step-by-step solution →
Q73·MathematicsNumerical
lim⁡x→23x+33−x−123−x/2−31−x\lim\limits_{x \to 2} \frac{3^{x}+3^{3-x}-12}{3^{-x/2}-3^{1-x}}x→2lim​3−x/2−31−x3x+33−x−12​ is equal to _________

Correct answer: 36

Step-by-step solution →
Q74·MathematicsNumerical
Let A(1,0),B(6,2)A(1,0), B(6,2)A(1,0),B(6,2) and C(32,6)C\left(\frac{3}{2}, 6\right)C(23​,6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point (−76,−13)\left(-\frac{7}{6}, -\frac{1}{3}\right)(−67​,−31​) is _________

Correct answer: 5

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Chemical Thermodynamics 165/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • Hydrogen 81/186
  • Carboxylic Acids and Derivatives 54/186
  • Diazonium Salts and Reactions 53/186
  • States of Matter: Gases and Liquids 52/186
  • Isomerism 51/186
  • Reaction Mechanism 29/186
  • Mathematical Reasoning 26/186
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