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JEE Main 9 January 2020 Shift 2 Question Paper with Answers

9 January 2020 · January session · 59 questions

59 of the 75 questions from the JEE Main 9 January 2020 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

16 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
19
Chemistry
20
Mathematics
20

Physics — JEE Main 9 January 2020 Shift 2

Q1·PhysicsSingle correct
A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):
  1. (A)r=3T(2d−ρ)gr = \sqrt{\frac{3T}{(2d - \rho)g}}r=(2d−ρ)g3T​​
  2. (B)r=T(d−ρ)gr = \sqrt{\frac{T}{(d - \rho)g}}r=(d−ρ)gT​​
  3. (C)r=2T3(d+ρ)gr = \sqrt{\frac{2T}{3(d + \rho)g}}r=3(d+ρ)g2T​​
  4. (D)r=T(d+ρ)gr = \sqrt{\frac{T}{(d + \rho)g}}r=(d+ρ)gT​​

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
There is a small source of light at some depth below the surface of water (refractive index = 43\frac{4}{3}34​) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly): [Use the fact that surface area of spherical cap of height h and radius of curvature r is 2πrh]
  1. (A)21%
  2. (B)17%
  3. (C)50%
  4. (D)34%

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A plane electromagnetic wave is propagating along the direction i^+j^2\frac{\hat{i} + \hat{j}}{\sqrt{2}}2​i^+j^​​, with its polarization along the direction k^\hat{k}k^. The correct form of the magnetic field of the wave would be (here B0B_{0}B0​ is an appropriate constant)
  1. (A)B0j^−i^2cos⁡(ωt+ki^+j^2)B_{0}\frac{\hat{j} - \hat{i}}{\sqrt{2}}\cos\left(\omega t + k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right)B0​2​j^​−i^​cos(ωt+k2​i^+j^​​)
  2. (B)B0 k^cos⁡(ωt−ki^+j^2)B_{0}\,\hat{k}\cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right)B0​k^cos(ωt−k2​i^+j^​​)
  3. (C)B0i^−j^2cos⁡(ωt−ki^+j^2)B_{0}\frac{\hat{i} - \hat{j}}{\sqrt{2}}\cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right)B0​2​i^−j^​​cos(ωt−k2​i^+j^​​)
  4. (D)B0i^+j^2cos⁡(ωt−ki^+j^2)B_{0}\frac{\hat{i} + \hat{j}}{\sqrt{2}}\cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right)B0​2​i^+j^​​cos(ωt−k2​i^+j^​​)

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
A wire of length L and mass per unit length 6.0×10−36.0 \times 10^{-3}6.0×10−3 kg m−1\mathrm{m}^{-1}m−1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is
  1. (A)1.1 m
  2. (B)5.1 m
  3. (C)2.1 m
  4. (D)8.1 m

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A particle of mass m is projected with a speed u form the ground at an angle θ=π3\theta = \frac{\pi}{3}θ=3π​ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity ui^u\hat{i}ui^. The horizontal distance covered by the combined mass before reaching the ground is:
  1. (A)58u2g\frac{5}{8}\frac{u^{2}}{g}85​gu2​
  2. (B)324u2g\frac{3\sqrt{2}}{4}\frac{u^{2}}{g}432​​gu2​
  3. (C)338u2g\frac{3\sqrt{3}}{8}\frac{u^{2}}{g}833​​gu2​
  4. (D)22u2g2\sqrt{2}\frac{u^{2}}{g}22​gu2​

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
In LC circuit the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V(t) = 10 sin (314 t) is applied to the circuit, the current in the circuit is given as
  1. (A)5.2 cos 314 t
  2. (B)0.52 sin 314 t
  3. (C)0.52 cos 314 t
  4. (D)10 cos 314 t

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
The current i in the network is:
  1. (A)0.6 A
  2. (B)0 A
  3. (C)0.2 A
  4. (D)0.3 A

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
A rod of length L has non-uniform linear mass density given by ρ(x)=a+b(xL)2\rho(x) = a + b\left(\frac{x}{L}\right)^{2}ρ(x)=a+b(Lx​)2, where a and b are constants and 0≤x≤L0 \le x \le L0≤x≤L. The value of x for the centre of mass of the rod is at:
  1. (A)32(a+b2a+b)L\frac{3}{2}\left(\frac{a + b}{2a + b}\right)L23​(2a+ba+b​)L
  2. (B)43(a+b2a+3b)L\frac{4}{3}\left(\frac{a + b}{2a + 3b}\right)L34​(2a+3ba+b​)L
  3. (C)34(2a+b3a+b)L\frac{3}{4}\left(\frac{2a + b}{3a + b}\right)L43​(3a+b2a+b​)L
  4. (D)32(2a+b3a+b)L\frac{3}{2}\left(\frac{2a + b}{3a + b}\right)L23​(3a+b2a+b​)L

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is
  1. (A)1:21 : \sqrt{2}1:2​
  2. (B)2:12 : 12:1
  3. (C)2:1\sqrt{2} : 12​:1
  4. (D)1:21 : 21:2

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
Planet A has mass M and radius R. Plant B has half the mass and half the radius of Planet A. If the escape velocities from the planets A and B are vAv_{A}vA​ and vBv_{B}vB​, respectively, then vAvB=n4\frac{v_{A}}{v_{B}} = \frac{n}{4}vB​vA​​=4n​. The value of ‘n’ is:
  1. (A)2
  2. (B)1
  3. (C)4
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
The energy required to ionize a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?
  1. (A)11.4 nm
  2. (B)35.8 nm
  3. (C)24.2 nm
  4. (D)8.6 nm

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
An electron of mass m and magnitude of charge ∣e∣|e|∣e∣ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is
  1. (A)∣e∣Eth\frac{|e|Et}{h}h∣e∣Et​
  2. (B)−h∣e∣Et-\frac{h}{|e|Et}−∣e∣Eth​
  3. (C)−h∣e∣Et-\frac{h}{|e|E\sqrt{t}}−∣e∣Et​h​
  4. (D)−h∣e∣Et2-\frac{h}{|e|Et^{2}}−∣e∣Et2h​

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
A particle starts from the origin at t = 0 with an initial velocity of 3.0 i^\hat{i}i^ m/s and moves in the x-y plane with a constant acceleration (6.0i^+4.0j^)(6.0\hat{i}+4.0\hat{j})(6.0i^+4.0j^​) m/s2^{2}2. The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is
  1. (A)40
  2. (B)60
  3. (C)32
  4. (D)50

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then:
  1. (A)T=πmIBT=\sqrt{\frac{\pi m}{IB}}T=IBπm​​
  2. (B)T=πm2IBT=\sqrt{\frac{\pi m}{2IB}}T=2IBπm​​
  3. (C)T=2πmIBT=\sqrt{\frac{2\pi m}{IB}}T=IB2πm​​
  4. (D)T=2mIBT=\sqrt{\frac{2m}{IB}}T=IB2m​​

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns/length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed v. If the electron does not hit the surface of the solenoid, maximum possible value of v is (all symbols have their standard meaning):
  1. (A)2eμ0nIRm\frac{2e\mu_{0}nIR}{m}m2eμ0​nIR​
  2. (B)eμ0nIRm\frac{e\mu_{0}nIR}{m}meμ0​nIR​
  3. (C)eμ0nIR4m\frac{e\mu_{0}nIR}{4m}4meμ0​nIR​
  4. (D)eμ0nIR2m\frac{e\mu_{0}nIR}{2m}2meμ0​nIR​

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
A spring mass system (mass m, spring constant k and natural length ℓ\ellℓ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity ω\omegaω, (k>>mω2)(k >> m\omega^{2})(k>>mω2) the relative change in the length of the spring is best given by the option:
  1. (A)mω23k\frac{m\omega^{2}}{3k}3kmω2​
  2. (B)mω2k\frac{m\omega^{2}}{k}kmω2​
  3. (C)2mω23k\frac{2m\omega^{2}}{3k}3k2mω2​
  4. (D)23(mω2k)\sqrt{\frac{2}{3}}\left(\frac{m\omega^{2}}{k}\right)32​​(kmω2​)

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsNumerical
In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is ℓ\ellℓ = 25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance ℓ′\ell'ℓ′ (in cm) will now be __________.

Correct answer: 40

Step-by-step solution →
Q18·PhysicsNumerical
An electric field E⃗=4xi^−(y2+1)j^\vec{E}=4x\hat{i}-(y^{2}+1)\hat{j}E=4xi^−(y2+1)j^​ N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI\phi_{I}ϕI​ and ϕII\phi_{II}ϕII​ respectively. The difference between (ϕI−ϕII\phi_{I}-\phi_{II}ϕI​−ϕII​) is (in N m2^{2}2/C) __________.

Correct answer: -48

Step-by-step solution →
Q19·PhysicsNumerical
In a Young's double's slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength λ\lambdaλ is used. Then the value of λ\lambdaλ is (in nm) __________.

Correct answer: 750

Step-by-step solution →

Chemistry — JEE Main 9 January 2020 Shift 2

Q20·ChemistrySingle correct
Amongst the following, the form of water with the lowest ionic conductance at 298 K is :
  1. (A)sea water
  2. (B)distilled water
  3. (C)saline water used for intravenous injection
  4. (D)water from a well

Correct answer: (B)

Step-by-step solution →
Q21·ChemistrySingle correct
Which polymer has 'chiral' monomer(s)?
  1. (A)Nylon 6, 6
  2. (B)Neoprene
  3. (C)PHBV
  4. (D)Buna-N

Correct answer: (C)

Step-by-step solution →
Q22·ChemistrySingle correct
The first and second ionisation enthalpies of a metal are 496 and 4560 kJ mol−1^{-1}−1, respectively. How many moles of HCl and H2_22​SO4_44​, respectively, will be needed to react completely with 1 mole of the metal hydroxide ?
  1. (A)1 and 1
  2. (B)1 and 2
  3. (C)2 and 0.5
  4. (D)1 and 0.5

Correct answer: (D)

Step-by-step solution →
Q23·ChemistrySingle correct
Among the statements (a) − (d), the correct ones are: (a) Lithium has the highest hydration enthalpy among the alkali metals. (b) Lithium chloride is insoluble in pyridine. (c) Lithium cannot form ethynide upon its reaction with ethyne (d) Both lithium and magnesium react slowly with H2_22​O
  1. (A)(a), (b) and (d) only
  2. (B)(a) and (d) only
  3. (C)(b) and (c) only
  4. (D)(a), (c) and (d) only

Correct answer: (D)

Step-by-step solution →
Q24·ChemistrySingle correct
The solubility product of Cr(OH)3_33​ at 298 K is 6.0×10−316.0 \times 10^{-31}6.0×10−31. The concentration of hydroxide ions in a saturated solution of Cr(OH)3_33​ will be :
  1. (A)(18×10−31)1/4(18 \times 10^{-31})^{1/4}(18×10−31)1/4
  2. (B)(4.86×10−29)1/4(4.86 \times 10^{-29})^{1/4}(4.86×10−29)1/4
  3. (C)(18×10−31)1/2(18 \times 10^{-31})^{1/2}(18×10−31)1/2
  4. (D)(2.22×10−31)1/4(2.22 \times 10^{-31})^{1/4}(2.22×10−31)1/4

Correct answer: (A)

Step-by-step solution →
Q25·ChemistrySingle correct
5 g of zinc is treated separately with an excess of (a) dilute hydrochloric acid and (b) aqueous sodium hydroxide. The ratio of the volumes of H2_22​ evolved in these two reactions is : -
  1. (A)1 : 4
  2. (B)1 : 1
  3. (C)1 : 2
  4. (D)2 : 1

Correct answer: (B)

Step-by-step solution →
Q26·ChemistrySingle correct
A mixture of gases O2_22​, H2_22​ and CO are taken in a closed vessel containing charcoal. The graph that represents the correct behaviour of pressure with time is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
The number of sp2^22 hybrid orbitals in a molecule of benzene is
  1. (A)18
  2. (B)6
  3. (C)12
  4. (D)24

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
Biochemical Oxygen Demand (BOD) is the amount of oxygen required (in ppm):
  1. (A)for the photochemical breakdown of waste present in 1 m3^33 volume of a water body.
  2. (B)for sustaining life in a water body.
  3. (C)by bacteria to break-down organic waste in a certain volume of a water sample.
  4. (D)by anaerobic bacteria to breakdown inorganic waste present in a water body.

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
A, B and C are three biomolecules. The results of the tests performed on them are given below: A, B and C are respectively:
Molisch's TestBarfoed TestBiuret Test
(A)PositiveNegativeNegative
(B)PositivePositiveNegative
(C)NegativeNegativePositive
  1. (A)A = Lactose, B = Glucose, C = Alanine
  2. (B)A = Lactose, B = Glucose, C = Albumin
  3. (C)A = Glucose, B = Fructose, C = Albumin
  4. (D)A = Lactose, B = Fructose, C = Alanine

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correct
Consider the following reactions: The compound[P] is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
Which of the following reactions will not produce a racemic product?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
Which of the following has the shortest C–Cl bond?
  1. (A)Cl – CH = CH – NO2_22​
  2. (B)Cl – CH = CH – OCH3_33​
  3. (C)Cl – CH = CH – CH3_33​
  4. (D)Cl – CH = CH2_22​

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
The decreasing order of basicity of the following amines is
  1. (A)(III) > (I) > (II) > (IV)
  2. (B)(II) > (III) > (IV) > (I)
  3. (C)(I) > (III) > (IV) > (II)
  4. (D)(III) > (II) > (I) > (IV)

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
The reaction of H3_33​N3_33​B3_33​Cl3_33​(A) with LiBH4_44​ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3_33​N3_33​B3_33​(Me)3_33​. Compounds (B) and (C) respectively, are:
  1. (A)Borazine and MeBr
  2. (B)Boron nitride and MeBr
  3. (C)Diborane and MeMgBr
  4. (D)Borazine and MeMgBr

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
In the following reaction A is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
The isomer (s) of [Co(NH3_33​)4_44​Cl2_22​] that has/have a Cl – Co – Cl angle of 90º, is/are :
  1. (A)meridional and trans
  2. (B)cis only
  3. (C)trans only
  4. (D)cis and trans

Correct answer: (B)

Step-by-step solution →
Q37·ChemistryNumerical
The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is ________________

Correct answer: 12

Step-by-step solution →
Q38·ChemistryNumerical
Consider the following reactions The mass percentage of carbon in A is____.

Correct answer: 66.67

Step-by-step solution →
Q39·ChemistryNumerical
10.30 mg of O2_22​ is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O2_22​ in ppm is________________.

Correct answer: 10

Step-by-step solution →

Mathematics — JEE Main 9 January 2020 Shift 2

Q40·MathematicsSingle correct
If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n=0}^{\infty} (-1)^{n} \tan^{2n}\thetax=n=0∑∞​(−1)ntan2nθ and y=∑n=0∞cos⁡2nθy = \sum\limits_{n=0}^{\infty} \cos^{2n}\thetay=n=0∑∞​cos2nθ, for 0<θ<π40 < \theta < \dfrac{\pi}{4}0<θ<4π​, then:
  1. (A)y(1−x)=1y(1 - x) = 1y(1−x)=1
  2. (B)y(1+x)=1y(1 + x) = 1y(1+x)=1
  3. (C)x(1+y)=1x(1 + y) = 1x(1+y)=1
  4. (D)x(1−y)=1x(1 - y) = 1x(1−y)=1

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correct
If p→(p∧∼q)p \rightarrow (p \wedge \sim q)p→(p∧∼q) is false, then the truth values of p and q are respectively:
  1. (A)F, T
  2. (B)F, F
  3. (C)T, T
  4. (D)T, F

Correct answer: (C)

Step-by-step solution →
Q42·MathematicsSingle correct
Let a,b∈R,a≠0a, b \in R, a \neq 0a,b∈R,a=0 be such that the equation ax2−2bx+5=0ax^{2} - 2bx + 5 = 0ax2−2bx+5=0 has a repeated root α\alphaα, which is also a root of the equation, x2−2bx−10=0x^{2} - 2bx - 10 = 0x2−2bx−10=0. If β\betaβ is the other root of this equation, then α2+β2\alpha^{2} + \beta^{2}α2+β2 is equal to:
  1. (A)28
  2. (B)24
  3. (C)26
  4. (D)25

Correct answer: (D)

Step-by-step solution →
Q43·MathematicsSingle correct
Let [t][t][t] denote the greatest integer ≤t\leq t≤t and lim⁡x→0x[4x]=A\lim\limits_{x \rightarrow 0} x\left[\dfrac{4}{x}\right] = Ax→0lim​x[x4​]=A. Then the function, f(x)=[x2]sin⁡(πx)f(x) = \left[x^{2}\right]\sin(\pi x)f(x)=[x2]sin(πx) is discontinuous, when x is equal to:
  1. (A)A+21\sqrt{A + 21}A+21​
  2. (B)A+5\sqrt{A + 5}A+5​
  3. (C)A\sqrt{A}A​
  4. (D)A+1\sqrt{A + 1}A+1​

Correct answer: (D)

Step-by-step solution →
Q44·MathematicsSingle correct
Let a function f:[0,5]→Rf : [0, 5] \rightarrow Rf:[0,5]→R be continuous, f(1)=3f(1) = 3f(1)=3 and F be defined as: F(x)=∫1xt2g(t) dtF(x) = \int\limits_{1}^{x} t^{2} g(t)\,dtF(x)=1∫x​t2g(t)dt, where g(t)=∫1tf(u) dug(t) = \int\limits_{1}^{t} f(u)\,dug(t)=1∫t​f(u)du. Then for the function F, the point x = 1 is:
  1. (A)a point of local minima
  2. (B)a point of inflection.
  3. (C)not a critical point
  4. (D)a point of local maxima

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correct
A random variable X has the following probability distribution: Then P(X>2)P(X > 2)P(X>2) is equal to:
X12345
P(X)K2K^{2}K22KK2K5K25K^{2}5K2
  1. (A)136\dfrac{1}{36}361​
  2. (B)712\dfrac{7}{12}127​
  3. (C)2336\dfrac{23}{36}3623​
  4. (D)16\dfrac{1}{6}61​

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correct
If ∫dθcos⁡2θ (tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int \dfrac{d\theta}{\cos^{2}\theta\,(\tan 2\theta + \sec 2\theta)} = \lambda \tan\theta + 2\log_{e}\left|f(\theta)\right| + C∫cos2θ(tan2θ+sec2θ)dθ​=λtanθ+2loge​∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ, f(θ))\left(\lambda,\, f(\theta)\right)(λ,f(θ)) is equal to:
  1. (A)(−1,1+tan⁡θ)(-1, 1 + \tan\theta)(−1,1+tanθ)
  2. (B)(1,1−tan⁡θ)(1, 1 - \tan\theta)(1,1−tanθ)
  3. (C)(1,1+tan⁡θ)(1, 1 + \tan\theta)(1,1+tanθ)
  4. (D)(−1,1−tan⁡θ)(-1, 1 - \tan\theta)(−1,1−tanθ)

Correct answer: (A)

Step-by-step solution →
Q47·MathematicsSingle correct
If dydx=xyx2+y2\dfrac{dy}{dx} = \dfrac{xy}{x^{2} + y^{2}}dxdy​=x2+y2xy​; y(1)=1y(1) = 1y(1)=1; then a value of x satisfying y(x)=ey(x) = ey(x)=e is:
  1. (A)2 e\sqrt{2}\,e2​e
  2. (B)123 e\dfrac{1}{2}\sqrt{3}\,e21​3​e
  3. (C)3 e\sqrt{3}\,e3​e
  4. (D)e2\dfrac{e}{\sqrt{2}}2​e​

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsSingle correct
Let ana_{n}an​ be the nthn^{th}nth term of a G.P. of positive terms. If ∑n=1100a2n+1=200\sum\limits_{n=1}^{100} a_{2n+1} = 200n=1∑100​a2n+1​=200 and ∑n=1100a2n=100\sum\limits_{n=1}^{100} a_{2n} = 100n=1∑100​a2n​=100, then ∑n=1200an\sum\limits_{n=1}^{200} a_{n}n=1∑200​an​ is equal to:
  1. (A)225
  2. (B)300
  3. (C)150
  4. (D)175

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correct
If z be a complex number satisfying ∣Re(z)∣+∣Im(z)∣=4\left|Re(z)\right| + \left|Im(z)\right| = 4∣Re(z)∣+∣Im(z)∣=4, then ∣z∣\left|z\right|∣z∣ cannot be:
  1. (A)172\sqrt{\dfrac{17}{2}}217​​
  2. (B)10\sqrt{10}10​
  3. (C)8\sqrt{8}8​
  4. (D)7\sqrt{7}7​

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
Given : f(x)={x,0≤x<1212,x=121−x,12<x≤1f(x) = \begin{cases} x, & 0 \leq x < \dfrac{1}{2} \\[4pt] \dfrac{1}{2}, & x = \dfrac{1}{2} \\[4pt] 1 - x, & \dfrac{1}{2} < x \leq 1 \end{cases}f(x)=⎩⎨⎧​x,21​,1−x,​0≤x<21​x=21​21​<x≤1​ and g(x)=(x−12)2,x∈Rg(x) = \left(x - \dfrac{1}{2}\right)^{2}, x \in Rg(x)=(x−21​)2,x∈R. Then the area (in sq. units) of the region bounded by the curves, y=f(x)y = f(x)y=f(x) and y=g(x)y = g(x)y=g(x) between the lines 2x=12x = 12x=1 and 2x=32x = \sqrt{3}2x=3​, is:
  1. (A)13+34\dfrac{1}{3} + \dfrac{\sqrt{3}}{4}31​+43​​
  2. (B)12+34\dfrac{1}{2} + \dfrac{\sqrt{3}}{4}21​+43​​
  3. (C)34−13\dfrac{\sqrt{3}}{4} - \dfrac{1}{3}43​​−31​
  4. (D)12−34\dfrac{1}{2} - \dfrac{\sqrt{3}}{4}21​−43​​

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
If A={x∈R:∣x∣<2}A = \{x \in R : |x| < 2\}A={x∈R:∣x∣<2} and B={x∈R:∣x−2∣≥3}B = \{x \in R : |x - 2| \ge 3\}B={x∈R:∣x−2∣≥3}; then:
  1. (A)B−A=R−(−2,5)B - A = R - (-2, 5)B−A=R−(−2,5)
  2. (B)A∩B=(−2,−1)A \cap B = (-2, -1)A∩B=(−2,−1)
  3. (C)A−B=[−1,2)A - B = [-1, 2)A−B=[−1,2)
  4. (D)A∪B=R−(2,5)A \cup B = R - (2, 5)A∪B=R−(2,5)

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correct
Let a−2b+c=1a - 2b + c = 1a−2b+c=1. If f(x)=∣x+ax+2x+1x+bx+3x+2x+cx+4x+3∣f(x) = \begin{vmatrix} x + a & x + 2 & x + 1 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix}f(x)=​x+ax+bx+c​x+2x+3x+4​x+1x+2x+3​​, then:
  1. (A)f(−50)=501f(-50) = 501f(−50)=501
  2. (B)f(50)=1f(50) = 1f(50)=1
  3. (C)f(50)=−501f(50) = -501f(50)=−501
  4. (D)f(−50)=−1f(-50) = -1f(−50)=−1

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
Let f and g be differentiable functions on R such that fog is the identity function. If for some a,b∈R,g′(a)=5a, b \in R, g'(a) = 5a,b∈R,g′(a)=5 and g(a)=bg(a) = bg(a)=b, then f′(b)f'(b)f′(b) is equal to:
  1. (A)111
  2. (B)555
  3. (C)15\frac{1}{5}51​
  4. (D)25\frac{2}{5}52​

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
If one end of a focal chord AB of the parabola y2=8xy^{2} = 8xy2=8x is at A(12,−2)A\left(\frac{1}{2}, -2\right)A(21​,−2), then the equation of the tangent to it at B is:
  1. (A)2x−y−24=02x - y - 24 = 02x−y−24=0
  2. (B)x−2y+8=0x - 2y + 8 = 0x−2y+8=0
  3. (C)2x+y−24=02x + y - 24 = 02x+y−24=0
  4. (D)x+2y+8=0x + 2y + 8 = 0x+2y+8=0

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
The length of the minor axis (along y − axis) of an ellipse in the standard form is 43\frac{4}{\sqrt{3}}3​4​. If this ellipse touches the line, x+6y=8x + 6y = 8x+6y=8; then its eccentricity is:
  1. (A)13113\frac{1}{3}\sqrt{\frac{11}{3}}31​311​​
  2. (B)1253\frac{1}{2}\sqrt{\frac{5}{3}}21​35​​
  3. (C)12113\frac{1}{2}\sqrt{\frac{11}{3}}21​311​​
  4. (D)56\sqrt{\frac{5}{6}}65​​

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsNumerical
If the curves, x2−6x+y2+8=0x^{2} - 6x + y^{2} + 8 = 0x2−6x+y2+8=0 and x2−8y+y2+16−k=0x^{2} - 8y + y^{2} + 16 - k = 0x2−8y+y2+16−k=0, (k>0)(k > 0)(k>0) touch each other at a point, then the largest value of k is ______.

Correct answer: 36

Step-by-step solution →
Q57·MathematicsNumerical
If Cr=25CrC_{r} = {}^{25}C_{r}Cr​=25Cr​ and C0+5.C1+9.C2+..........+(101).C25=225.kC_{0} + 5.C_{1} + 9.C_{2} + .......... + (101).C_{25} = 2^{25}.kC0​+5.C1​+9.C2​+..........+(101).C25​=225.k, then k is equal to ______

Correct answer: 51

Step-by-step solution →
Q58·MathematicsNumerical
Let a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c be three vectors such that ∣a⃗∣=3,∣b⃗∣=5,b⃗.c⃗=10|\vec{a}| = \sqrt{3}, |\vec{b}| = 5, \vec{b}.\vec{c} = 10∣a∣=3​,∣b∣=5,b.c=10 and the angle between b⃗\vec{b}b and c⃗\vec{c}c is π3\frac{\pi}{3}3π​. If a⃗\vec{a}a is perpendicular to the vector b⃗×c⃗\vec{b} \times \vec{c}b×c, then ∣a⃗×(b⃗×c⃗)∣|\vec{a} \times (\vec{b} \times \vec{c})|∣a×(b×c)∣ is equal to ____.

Correct answer: 30

Step-by-step solution →
Q59·MathematicsNumerical
The number of terms common to the two A.P.'s 3, 7, 11,…………..407 and 2, 9, 16, …..709 is ______

Correct answer: 14

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
  • 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
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • 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
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Laws of Motion 130/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Circles 142/186
  • 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
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alternating Currents 108/186
  • Waves 109/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Ellipse 103/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
  • Indefinite Integration 66/186
  • Polymers 64/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
  • Mathematical Reasoning 26/186
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