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JEE Main 3 September 2020 Shift 1 Question Paper with Answers

3 September 2020 · September session · 72 questions

72 of the 75 questions from the JEE Main 3 September 2020 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 questions are 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
23
Mathematics
24

Physics — JEE Main 3 September 2020 Shift 1

Q1·PhysicsSingle correct
A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wave-train of wavelength 6 cm is produced at the lower and the rope. What is the wavelength of the wave train (in cm) when it reaches the top of the tope?
  1. (A)6
  2. (B)12
  3. (C)3
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
When the wavelength of radiation falling on a metal is charged from 500 nm to 200 nm the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metals is close to:
  1. (A)1.02 eV
  2. (B)0.61 eV
  3. (C)0.52 eV
  4. (D)0.81 eV

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to its as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length ℓ\ellℓ = 1 m, the value of θ is approximately: (take g = 10 m/s2^{2}2)
  1. (A)49°
  2. (B)55°
  3. (C)69°
  4. (D)63°

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
An elliptical loop having resistance R, of semi major axis a and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ω, the average power loss in the loop due to joule heating is:
  1. (A)π2a2b2B2ω2R\frac{\pi^{2}a^{2}b^{2}B^{2}\omega^{2}}{R}Rπ2a2b2B2ω2​
  2. (B)zero
  3. (C)πabBωR\frac{\pi abB\omega}{R}RπabBω​
  4. (D)π2a2b2B2ω22R\frac{\pi^{2}a^{2}b^{2}B^{2}\omega^{2}}{2R}2Rπ2a2b2B2ω2​

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
Model a torch battery of length ℓ\ellℓ to be made up of a thin cylindrical bar of radius 'a' and a concentric thin cylindrical shell of radius 'b' filled in between with an electrolyte of resistivity ρ (see figure). If the battery is connected to a resistance of value R, the maximum joule heating in R will take place for:
  1. (A)R=ρπℓℓn(ba)R = \frac{\rho}{\pi\ell}\ell n\left(\frac{b}{a}\right)R=πℓρ​ℓn(ab​)
  2. (B)R=ρ2πℓ(ba)R = \frac{\rho}{2\pi\ell}\left(\frac{b}{a}\right)R=2πℓρ​(ab​)
  3. (C)R=2ρπℓℓn(ba)R = \frac{2\rho}{\pi\ell}\ell n\left(\frac{b}{a}\right)R=πℓ2ρ​ℓn(ab​)
  4. (D)R=ρ2πℓℓn(ba)R = \frac{\rho}{2\pi\ell}\ell n\left(\frac{b}{a}\right)R=2πℓρ​ℓn(ab​)

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is
  1. (A)35\frac{3}{5}53​
  2. (B)34\frac{3}{4}43​
  3. (C)78\frac{7}{8}87​
  4. (D)45\frac{4}{5}54​

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
A charged particle carrying charge 1 μC is moving with velocity (2i^+3j^+4jk^)\left(2\hat{i} + 3\hat{j} + 4j\hat{k}\right)(2i^+3j^​+4jk^) ms−1^{-1}−1. If an external magnetic field of (5i^+3j^−6jk^)×10−3\left(5\hat{i} + 3\hat{j} - 6j\hat{k}\right) \times 10^{-3}(5i^+3j^​−6jk^)×10−3 T exists in the region where the particle is moving then the force on the particle is F⃗×10−9\vec{F} \times 10^{-9}F×10−9 N. The vector F⃗\vec{F}F is:
  1. (A)−30i^+32j^−9k^-30\hat{i} + 32\hat{j} - 9\hat{k}−30i^+32j^​−9k^
  2. (B)−3.0i^+3.2j^−0.9k^-3.0\hat{i} + 3.2\hat{j} - 0.9\hat{k}−3.0i^+3.2j^​−0.9k^
  3. (C)−300i^+320j^−90k^-300\hat{i} + 320\hat{j} - 90\hat{k}−300i^+320j^​−90k^
  4. (D)−0.30i^+0.32j^−0.09k^-0.30\hat{i} + 0.32\hat{j} - 0.09\hat{k}−0.30i^+0.32j^​−0.09k^

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:
  1. (A)1.7°
  2. (B)0.07°
  3. (C)0.57°
  4. (D)0.17°

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
In the circuit shown in the figure, the total charge is 750 μC and the voltage across capacitor C2_{2}2​ is 20 V. Then the charge on capacitor C2_{2}2​ is
  1. (A)160 μC
  2. (B)650 μC
  3. (C)590 μC
  4. (D)450 μC

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω an inductance of 0.1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to:
  1. (A)348 s
  2. (B)365 s
  3. (C)418 s
  4. (D)245 s

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is
  1. (A)4 : 1
  2. (B)0.8 : 1
  3. (C)2 : 1
  4. (D)8 : 1

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius ReR_eRe​. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become 32\sqrt{\dfrac{3}{2}}23​​ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is:
  1. (A)2.5Re2.5R_e2.5Re​
  2. (B)3Re3R_e3Re​
  3. (C)2Re2R_e2Re​
  4. (D)4Re4R_e4Re​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm. 50 turns and carrying current I (Ampere) in units of μ0Iπ\dfrac{\mu_0 I}{\pi}πμ0​I​ is:
  1. (A)5003500\sqrt{3}5003​
  2. (B)50350\sqrt{3}503​
  3. (C)535\sqrt{3}53​
  4. (D)2503250\sqrt{3}2503​

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
The magnetic field of a plane electromagnetic wave is B⃗=3×10−8 sin⁡[200 π(y+ct)] i^ T\vec{B} = 3 \times 10^{-8}\ \sin[200\ \pi(y+ ct)]\ \hat{i}\,TB=3×10−8 sin[200 π(y+ct)] i^T where c=3×108 ms−1c = 3 \times 10^{8}\ ms^{-1}c=3×108 ms−1 is the speed of light. The corresponding electric field is
  1. (A)E⃗=9sin⁡[200 π(y+ct)] k^\vec{E} = 9\sin[200\ \pi(y+ ct)]\ \hat{k}E=9sin[200 π(y+ct)] k^ V / m
  2. (B)E⃗=−9sin⁡[200 π(y+ct)] k^\vec{E} = -9\sin[200\ \pi(y+ ct)]\ \hat{k}E=−9sin[200 π(y+ct)] k^ V / m
  3. (C)E⃗=3×10−8sin⁡[200 π(y+ct)] k^\vec{E} = 3\times 10^{-8}\sin[200\ \pi(y+ ct)]\ \hat{k}E=3×10−8sin[200 π(y+ct)] k^ V / m
  4. (D)E⃗=−10−6sin⁡[200 π(y+ct)] k^\vec{E} = -10^{-6}\sin[200\ \pi(y+ ct)]\ \hat{k}E=−10−6sin[200 π(y+ct)] k^ V / m

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
  1. (A)2.121 cm
  2. (B)2.123 cm
  3. (C)2.124 cm
  4. (D)2.125 cm

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
Consider a gas of triatomic molecules. The molecules are assumed to be triangular and made of massless rigid rods whose vertices are occupied by atoms. The internal energy of a mole of the gas at temperature T is:
  1. (A)52RT\dfrac{5}{2}RT25​RT
  2. (B)32RT\dfrac{3}{2}RT23​RT
  3. (C)3RT3RT3RT
  4. (D)92RT\dfrac{9}{2}RT29​RT

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Two isolated conducting spheres S1S_1S1​ and S2S_2S2​ of radius 23R\dfrac{2}{3}R32​R and 13R\dfrac{1}{3}R31​R have 12 μC and −3 μC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S1S_1S1​ and S2S_2S2​ are respectively:
  1. (A)+4.5 μC and −4.5 μC
  2. (B)4.5 μC on both
  3. (C)6 μC and 3 μC
  4. (D)3 μC and 6 μC

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is
  1. (A)100 Ω
  2. (B)200 Ω
  3. (C)50 Ω
  4. (D)300 Ω

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correct
Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is I=M(R24+L212)I = M\left(\dfrac{R^{2}}{4} + \dfrac{L^{2}}{12}\right)I=M(4R2​+12L2​). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is
  1. (A)23\dfrac{2}{3}32​
  2. (B)32\sqrt{\dfrac{3}{2}}23​​
  3. (C)32\dfrac{3}{2}23​
  4. (D)23\sqrt{\dfrac{2}{3}}32​​

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A balloon filled with helium (32°C and 1.7 atm.) bursts. Immediately afterwards the expansion of helium can be considered as:
  1. (A)irreversible adiabatic
  2. (B)reversible isothermal
  3. (C)reversible adiabatic
  4. (D)irreversible isothermal

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is N / 100, where N is an integer, the value of N is __________.

Correct answer: 158.00

Step-by-step solution →
Q22·PhysicsNumerical
A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre __________.

Correct answer: 9.00

Step-by-step solution →
Q23·PhysicsNumerical
When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to 0°, the surface tension of the liquid, in milli Newton m−1m^{-1}m−1, is [ρ(lliquid)\rho_{(lliquid)}ρ(lliquid)​ = 900 kgm−3kgm^{-3}kgm−3, g = 10 ms−2ms^{-2}ms−2] (Give answer in closest integer)

Correct answer: 101

Step-by-step solution →
Q24·PhysicsNumerical
A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms−2ms^{-2}ms−2) __________.

Correct answer: 150.00

Step-by-step solution →
Q25·PhysicsNumerical
A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with VmV_mVm​ volume (at 30°C) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If γ(beaker)\gamma_{(beaker)}γ(beaker)​ = 6 × 10−610^{-6}10−6 °C−1C^{-1}C−1 and γ(mercury)\gamma_{(mercury)}γ(mercury)​ = 1.5 × 10−410^{-4}10−4 °C−1C^{-1}C−1, where γ is the coefficient of volume expansion, then VmV_mVm​ (in cc) is close to __________.

Correct answer: 20.00

Step-by-step solution →

Chemistry — JEE Main 3 September 2020 Shift 1

Q26·ChemistrySingle correct
Aqua regia is used for dissolving noble metals (Au, Pt, etc.). The gas evolved in this process is:
  1. (A)NO
  2. (B)N2_22​O3_33​
  3. (C)N2_22​
  4. (D)N2_22​O5_55​

Correct answer: (A)

Step-by-step solution →
Q27·ChemistrySingle correct
An acidic buffer is obtained on mixing:
  1. (A)100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl
  2. (B)100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3_33​COONa
  3. (C)100 mL of 0.1 M CH3_33​COOH and 100 mL of 0.1 M NaOH
  4. (D)100 mL of 0.1 M CH3_33​COOH and 200 mL of 0.1 M NaOH

Correct answer: (B)

Step-by-step solution →
Q28·ChemistrySingle correct
The mechanism of SN_NN​1 reaction is given as: A student writes general characteristics based on the given mechanism as: (a) The reaction is favoured by weak nucleophiles. (b) R⊕^\oplus⊕ would be easily formed if the substituents are bulky. (c) The reaction is accompanied by racemization. (d) The reaction is favoured by non-polar solvents. Which observations are correct?
  1. (A)(b) and (d)
  2. (B)(a), (b) and (c)
  3. (C)(a) and (c)
  4. (D)(a) and (b)

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correct
Of the species, NO, NO+^++, NO2+^{2+}2+ and NO−^-−, the one with minimum bond strength is
  1. (A)NO+^++
  2. (B)NO
  3. (C)NO2+^{2+}2+
  4. (D)NO−^-−

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
Which one of the following compounds possesses the most acidic hydrogen?
  1. (A)H3_33​C −-− C ≡\equiv≡ C −-− H
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
The electronic spectrum of [Ti(H2_22​O)6_66​]3+^{3+}3+ shows a single broad peak with a maximum at 20,300 cm−1^{-1}−1. The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol−1^{-1}−1, is: (1 kJ mol−1^{-1}−1 = 83.7 cm−1^{-1}−1)
  1. (A)242.5
  2. (B)97
  3. (C)83.7
  4. (D)145.5

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
The complex that can shows optical activity is:
  1. (A)cis-[CrCl2_22​(ox)2_22​]3−^{3-}3− (ox = oxalate)
  2. (B)trans-[Cr(Cl2_22​)(ox)2_22​]3−^{3-}3−
  3. (C)cis-[Fe(NH3_33​)2_22​(CN)4_44​]−^-−
  4. (D)trans-[Fe(NH3_33​)2_22​(CN)4_44​]−^-−

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
Thermal power plants can lead to:
  1. (A)Ozone layer depletion
  2. (B)Acid rain
  3. (C)Blue baby syndrome
  4. (D)Eutrophication

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
In a molecule of pyrophosphoric acid, the number of P −-− OH, P === O and P −-− O −-− P bonds/ moiety(ies) respectively are
  1. (A)4, 2 and 1
  2. (B)4, 2 and 0
  3. (C)2, 4 and 1
  4. (D)3, 3 and 3

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
The atomic number of the element unnilennium is
  1. (A)102
  2. (B)108
  3. (C)119
  4. (D)109

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Glycerol is separated in soap industries by:
  1. (A)Differential extraction
  2. (B)Fractional distillation
  3. (C)Distillation under reduced pressure
  4. (D)Steam distillation

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
Henry's constant (in kbar) for four gases α, β, γ and δ in water at 298 K is given below: (density of water = 103^33 kg m−3^{-3}−3 at 298 K) This table implies that:
αβγδ
KH_HH​5022 ×\times× 10−5^{-5}−50.5
  1. (A)α has the highest solubility in water at a given pressure
  2. (B)solubility of γ at 308 K is lower than at 298 K.
  3. (C)The pressure of a 55.5 molal solution of δ is 250 bar.
  4. (D)The pressure of a 55.5 molal solution of γ is 1 bar.

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
The Khjeldahl method of Nitrogen estimation fails for which of the following reaction products?
  1. (A)(c) and (d)
  2. (B)(b) and (c)
  3. (C)(a) and (d)
  4. (D)(a), (c) and (d)

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
The antifertility drug "Noverstrol" can react with:
  1. (A)Alcoholic HCN; NaOCl; ZnCl2_22​/HCl
  2. (B)Br2_22​/water; ZnCl2_22​/ HCl; NaOCl
  3. (C)ZnCl2_22​/HCl ; FeCl3_33​; Alcoholic HCN
  4. (D)Br2_22​/water; ZnCl2_22​/HCl; FeCl3_33​

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
If the boiling point of H2_22​O is 373 K, the boiling point of H2_22​S will be:
  1. (A)more than 373 K
  2. (B)less than 300 K
  3. (C)greater than 300 K but less than 373 K
  4. (D)equal to 373 K

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
Tyndall effect is observed when
  1. (A)The diameter of dispersed particles is similar to the wavelength of light used.
  2. (B)The diameter of dispersed particles is much larger than the wavelength of light used.
  3. (C)The refractive index of dispersed phase is greater than that of the dispersion medium.
  4. (D)The diameter of dispersed particles is much smaller than the wavelength of light used.

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
It is true that:
  1. (A)A first order reaction is always a single step reaction.
  2. (B)A zero order reaction is a single step reaction.
  3. (C)A second order reaction is always a multistep reaction.
  4. (D)A zero order reaction is a multistep reaction.

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
Which of the following compounds produces an optically inactive compound on hydrogenation?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistryNumerical
The volume strength of 8.9 M H2_22​O2_22​ solution calculated at 273 K and 1 atm is __________. (R = 0.0821 L atm K−1^{-1}−1 mol−1^{-1}−1) (rounded off to the nearest integer)

Correct answer: 100

Step-by-step solution →
Q45·ChemistryNumerical
The photoelectric current from Na (work function, w0_00​ = 2.3 eV) is stopped by the output voltage of the cell Pt(s)/H2_22​(g, 1 bar) | HCl (aq., pH = 1) | AgCl(s) | Ag(s). The pH of aq. HCl required to stop the photoelectric current from K(w0_00​ = 2.25 eV), all other conditions remaining the same, is __________ ×\times× 10−2^{-2}−2 (to the nearest integer). Given, 2.303 RTF\frac{RT}{F}FRT​ = 0.06 V ; EAgCl ∣ Ag ∣ Cl−0^0_{AgCl\,|\,Ag\,|\,Cl^-}AgCl∣Ag∣Cl−0​ = 0.22 V

Correct answer: 142

Step-by-step solution →
Q46·ChemistryNumerical
The total number of monohalogenated organic products in the following (including stereoisomers) reaction is __________. A (Simplest optically active alkene) →(ii) X2/Δ(i) H2/Ni/Δ\xrightarrow[\text{(ii) X}_2/\Delta]{\text{(i) H}_2/\text{Ni}/\Delta}(i) H2​/Ni/Δ(ii) X2​/Δ​

Correct answer: 8

Step-by-step solution →
Q47·ChemistryNumerical
An element with molar mass 2.7 ×\times× 10−2^{-2}−2 kg mol−1^{-1}−1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 ×\times× 103^33 kg m−3^{-3}−3, the radius of the element is approximately __________ ×\times× 10−12^{-12}−12 m (to the nearest integer).

Correct answer: 143

Step-by-step solution →
Q48·ChemistryNumerical
The mole fraction of glucose (C6_66​H12_{12}12​O6_66​) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is __________.

Correct answer: 47

Step-by-step solution →

Mathematics — JEE Main 3 September 2020 Shift 1

Q49·MathematicsSingle correct
A hyperbola having the transverse axis of length 2\sqrt{2}2​ has the same foci as that of the ellipse 3x2+4y2=123x^{2}+4y^{2}=123x2+4y2=12, then this hyperbola does not pass through which of the following points?
  1. (A)(1,−12)\left(1,-\frac{1}{\sqrt{2}}\right)(1,−2​1​)
  2. (B)(32,12)\left(\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}}\right)(23​​,2​1​)
  3. (C)(−32,1)\left(-\sqrt{\frac{3}{2}},1\right)(−23​​,1)
  4. (D)(12,0)\left(\frac{1}{\sqrt{2}},0\right)(2​1​,0)

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correct
Consider the two sets: A = {m ∈ R : both the roots of x2−(m+1)x+m+4=0x^{2}-(m+1)x+m+4=0x2−(m+1)x+m+4=0 are real} and B = [−3, 5). Which of the following is not true?
  1. (A)A∩B={−3}A\cap B=\{-3\}A∩B={−3}
  2. (B)B−A=(−3, 5)B-A=(-3,\ 5)B−A=(−3, 5)
  3. (C)A−B=(−∞, −3)∪(5, ∞)A-B=(-\infty,\ -3)\cup(5,\ \infty)A−B=(−∞, −3)∪(5, ∞)
  4. (D)A∪B=RA\cup B=RA∪B=R

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
If y2+log⁡e(cos⁡2x)=yy^{2}+\log_{e}\left(\cos^{2}x\right)=yy2+loge​(cos2x)=y, x∈(−π2,π2)x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)x∈(−2π​,2π​), Then:
  1. (A)∣y′(0)∣+∣y′′(0)∣=3\left|y'(0)\right|+\left|y''(0)\right|=3∣y′(0)∣+∣y′′(0)∣=3
  2. (B)∣y′′(0)∣=2\left|y''(0)\right|=2∣y′′(0)∣=2
  3. (C)∣y′(0)∣+∣y′′(0)∣=1\left|y'(0)\right|+\left|y''(0)\right|=1∣y′(0)∣+∣y′′(0)∣=1
  4. (D)∣y′′(0)∣=0\left|y''(0)\right|=0∣y′′(0)∣=0

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
Let P be a point on the parabola, y2=12xy^{2}=12xy2=12x and N be the foot of the perpendicular drawn from P on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 43\frac{4}{3}34​, then:
  1. (A)MQ=13MQ=\frac{1}{3}MQ=31​
  2. (B)MQ=14MQ=\frac{1}{4}MQ=41​
  3. (C)PN=4PN=4PN=4
  4. (D)PN=3PN=3PN=3

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
The lines r⃗=(i^−j^)+ℓ(2i^+k^)\vec{r}=\left(\hat{i}-\hat{j}\right)+\ell\left(2\hat{i}+\hat{k}\right)r=(i^−j^​)+ℓ(2i^+k^) and r⃗=(2i^−j^)+m(i^+j^−k^)\vec{r}=\left(2\hat{i}-\hat{j}\right)+m\left(\hat{i}+\hat{j}-\hat{k}\right)r=(2i^−j^​)+m(i^+j^​−k^)
  1. (A)intersect when ℓ=2\ell=2ℓ=2 and m=12m=\frac{1}{2}m=21​
  2. (B)intersect when ℓ=1\ell=1ℓ=1 and m=2m=2m=2
  3. (C)do not intersect for any values of ℓ\ellℓ and mmm
  4. (D)intersect for all values of ℓ\ellℓ and mmm

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, −2, 3) and (1, 1, 0) lies on the plane:
  1. (A)x−y−2z=1x-y-2z=1x−y−2z=1
  2. (B)2x+y−z=12x+y-z=12x+y−z=1
  3. (C)x−2y+z=1x-2y+z=1x−2y+z=1
  4. (D)x+2y−z=1x+2y-z=1x+2y−z=1

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
The solution curve of the differential equation, (1+e−x)(1+y2)dydx=y2\left(1+e^{-x}\right)\left(1+y^{2}\right)\frac{dy}{dx}=y^{2}(1+e−x)(1+y2)dxdy​=y2, which passes through the point (0, 1), is
  1. (A)y2+1=y(log⁡e(1+ex2)+2)y^{2}+1=y\left(\log_{e}\left(\frac{1+e^{x}}{2}\right)+2\right)y2+1=y(loge​(21+ex​)+2)
  2. (B)y2+1=ylog⁡e(1+ex2)y^{2}+1=y\log_{e}\left(\frac{1+e^{x}}{2}\right)y2+1=yloge​(21+ex​)
  3. (C)y2+1=y(log⁡e(1+e−x2)+2)y^{2}+1=y\left(\log_{e}\left(\frac{1+e^{-x}}{2}\right)+2\right)y2+1=y(loge​(21+e−x​)+2)
  4. (D)y2=1+ylog⁡e(1+ex2)y^{2}=1+y\log_{e}\left(\frac{1+e^{x}}{2}\right)y2=1+yloge​(21+ex​)

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsSingle correct
For the frequency distribution: Variate (x): x1x_{1}x1​ x2x_{2}x2​ x3x_{3}x3​ ....x15x_{15}x15​ Frequency (f): f1f_{1}f1​ f2f_{2}f2​ f3f_{3}f3​ ....f15f_{15}f15​ Where 0<x1<x2<x3<…<x15=100<x_{1}<x_{2}<x_{3}<\ldots<x_{15}=100<x1​<x2​<x3​<…<x15​=10 and ∑i=115fi>0\sum\limits_{i=1}^{15}f_{i}>0i=1∑15​fi​>0, the standard deviation cannot be:
  1. (A)4
  2. (B)2
  3. (C)6
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q57·MathematicsSingle correct
If Δ=∣x−22x−33x−42x−33x−44x−53x−55x−810x−17∣=Ax3+Bx2+Cx+D\Delta=\begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ 2x-3 & 3x-4 & 4x-5 \\ 3x-5 & 5x-8 & 10x-17 \end{vmatrix}=Ax^{3}+Bx^{2}+Cx+DΔ=​x−22x−33x−5​2x−33x−45x−8​3x−44x−510x−17​​=Ax3+Bx2+Cx+D, then B + C is equal to:
  1. (A)−3
  2. (B)9
  3. (C)−1
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correct
The proposition p→∼(p∧∼q)p \rightarrow \sim (p \wedge \sim q)p→∼(p∧∼q) is equivalent to:
  1. (A)(∼p)∧q(\sim p) \wedge q(∼p)∧q
  2. (B)qqq
  3. (C)(∼p)∨q(\sim p) \vee q(∼p)∨q
  4. (D)(∼p)∨(∼q)(\sim p) \vee (\sim q)(∼p)∨(∼q)

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
∫−ππ∣π−∣x∣∣dx\int\limits_{-\pi}^{\pi} \left| \pi - |x| \right| dx−π∫π​∣π−∣x∣∣dx is equal to:
  1. (A)π2\pi^{2}π2
  2. (B)π22\frac{\pi^{2}}{2}2π2​
  3. (C)2 π2\sqrt{2}\,\pi^{2}2​π2
  4. (D)2π22\pi^{2}2π2

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
A die is thrown two times and the sum of the scorers appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is
  1. (A)19\frac{1}{9}91​
  2. (B)13\frac{1}{3}31​
  3. (C)18\frac{1}{8}81​
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is:
  1. (A)14\frac{1}{4}41​
  2. (B)17\frac{1}{7}71​
  3. (C)16\frac{1}{6}61​
  4. (D)15\frac{1}{5}51​

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
If the number of integral terms in the expansion of (312+518)n\left(3^{\frac{1}{2}} + 5^{\frac{1}{8}}\right)^{n}(321​+581​)n is exactly 33, then the least value of 'n' is
  1. (A)264
  2. (B)248
  3. (C)256
  4. (D)128

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
The function, f(x)=(3x−7)x2/3f(x) = (3x - 7)x^{2/3}f(x)=(3x−7)x2/3, x∈Rx \in Rx∈R, is increasing for all x lying in:
  1. (A)(−∞,1415)\left(-\infty, \frac{14}{15}\right)(−∞,1514​)
  2. (B)(−∞,0)∪(37,∞)(-\infty, 0) \cup \left(\frac{3}{7}, \infty\right)(−∞,0)∪(73​,∞)
  3. (C)(−∞,−1415)∪(0,∞)\left(-\infty, -\frac{14}{15}\right) \cup (0, \infty)(−∞,−1514​)∪(0,∞)
  4. (D)(−∞,0)∪(1415,∞)(-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)(−∞,0)∪(1514​,∞)

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
2π−(sin⁡−145+sin⁡−1513+sin⁡−11665)2\pi - \left( \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65} \right)2π−(sin−154​+sin−1135​+sin−16516​) is equal to:
  1. (A)3π2\frac{3\pi}{2}23π​
  2. (B)5π4\frac{5\pi}{4}45π​
  3. (C)7π4\frac{7\pi}{4}47π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
Let [t][t][t] denote the greatest integer ≤t\leq t≤t. If for some λ∈R−{0,1}\lambda \in R - \{0, 1\}λ∈R−{0,1}, lim⁡x→0∣1−x+∣x∣λ−x+[x]∣=L\lim\limits_{x \to 0} \left| \frac{1 - x + |x|}{\lambda - x + [x]} \right| = Lx→0lim​​λ−x+[x]1−x+∣x∣​​=L, then L is equal to:
  1. (A)0
  2. (B)1
  3. (C)12\frac{1}{2}21​
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
The area (in sq. units) of the region {(x,y):0≤y≤x2+1, 0≤y≤x+1, 12≤x≤2}\{(x, y) : 0 \leq y \leq x^{2} + 1,\ 0 \leq y \leq x + 1,\ \frac{1}{2} \leq x \leq 2\}{(x,y):0≤y≤x2+1, 0≤y≤x+1, 21​≤x≤2} is:
  1. (A)7916\frac{79}{16}1679​
  2. (B)2316\frac{23}{16}1623​
  3. (C)7924\frac{79}{24}2479​
  4. (D)236\frac{23}{6}623​

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
The value of (2⋅1P0−3⋅2P1+4⋅3P2−…(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \ldots(2⋅1P0​−3⋅2P1​+4⋅3P2​−… up to 51th51^{th}51th term)+(1!−2!+3!−…) + (1! - 2! + 3! - \ldots)+(1!−2!+3!−… upt to 51th51^{th}51th term))) is equal to:
  1. (A)1
  2. (B)1+(51)!1 + (51)!1+(51)!
  3. (C)1+(52)!1 + (52)!1+(52)!
  4. (D)1−51(51)!1 - 51(51)!1−51(51)!

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsNumerical
Let A=[x110]A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}A=[x1​10​], x∈Rx \in Rx∈R and A4=[aij]A^{4} = [a_{ij}]A4=[aij​]. If a11=109a_{11} = 109a11​=109, then a22a_{22}a22​ is equal to __________.

Correct answer: 10

Step-by-step solution →
Q69·MathematicsNumerical
The diameter of the circle, whose centre lies on the lines x+y=2x + y = 2x+y=2 in the first quadrant and which touches both the lines x=3x = 3x=3 and y=2y = 2y=2, is __________.

Correct answer: 3

Step-by-step solution →
Q70·MathematicsNumerical
If lim⁡x→0{1x8(1−cos⁡x22−cos⁡x24+cos⁡x22cos⁡x24)}=2−k\lim\limits_{x \to 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos\frac{x^{2}}{2} - \cos\frac{x^{2}}{4} + \cos\frac{x^{2}}{2}\cos\frac{x^{2}}{4} \right) \right\} = 2^{-k}x→0lim​{x81​(1−cos2x2​−cos4x2​+cos2x2​cos4x2​)}=2−k, then the value of k is __________.

Correct answer: 8

Step-by-step solution →
Q71·MathematicsNumerical
The value of (0.16)log⁡2.5(13+132+133+… to ∞)(0.16)^{\log_{2.5}\left(\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \ldots \text{ to } \infty\right)}(0.16)log2.5​(31​+321​+331​+… to ∞) is equal to __________.

Correct answer: 4

Step-by-step solution →
Q72·MathematicsNumerical
If (1+i1−i)m/2=(1+ii−1)n/3=1\left(\frac{1+i}{1-i}\right)^{m/2} = \left(\frac{1+i}{i-1}\right)^{n/3} = 1(1−i1+i​)m/2=(i−11+i​)n/3=1, (m,n∈N)(m, n \in N)(m,n∈N) Then the greatest common divisor of the least values of m and n is __________.

Correct answer: 4

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
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • 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
  • Complex Numbers 165/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Work, Energy and Power 132/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Differentiability 91/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • Isomerism 51/186
  • Reaction Mechanism 29/186
  • Mathematical Reasoning 26/186
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