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JEE Main 10 April 2019 Shift 1 Question Paper with Answers

10 April 2019 · April session · 77 questions

77 of the 90 questions from the JEE Main 10 April 2019 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

13 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
24
Chemistry
29
Mathematics
24

Physics — JEE Main 10 April 2019 Shift 1

Q1·PhysicsSingle correct
A particle of mass m is moving along a trajectory given by x=x0+acos⁡ω1tx = x_{0} + a\cos\omega_{1}tx=x0​+acosω1​t y=y0+bsin⁡ω2ty = y_{0} + b\sin\omega_{2}ty=y0​+bsinω2​t The torque, acing on the particle about the origin, at t = 0 is:
  1. (A)my0aω12k^my_{0}a\omega_{1}^{2}\hat{k}my0​aω12​k^
  2. (B)m(−x0b+y0a)ω12k^m(-x_{0}b + y_{0}a)\omega_{1}^{2}\hat{k}m(−x0​b+y0​a)ω12​k^
  3. (C)−m(−x0bω22−y0aω12)k^-m(-x_{0}b\omega_{2}^{2} - y_{0}a\omega_{1}^{2})\hat{k}−m(−x0​bω22​−y0​aω12​)k^
  4. (D)Zero

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in ms−1^{-1}−1 (Given speed of sound = 300 m/s)
  1. (A)16, 14
  2. (B)12, 16
  3. (C)8, 18
  4. (D)12, 18

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
Given below in the left column are different modes of communication using the kinds of waves given in the right column. From the options given below, find the most appropriate match between entries in the left and the right column.
Modes of CommunicationKinds of Waves
A.Optical Fibre CommunicationP.Ultrasound
B.RadarQ.Infrared Light
C.SonarR.Microwaves
D.Mobile PhonesS.Radio Waves
  1. (A)A - S, B - Q, C - R, D - P
  2. (B)A - Q, B - S, C - P, D - R
  3. (C)A - R, B - P, C - S, D - Q
  4. (D)A - Q, B - S, C - R, D - P

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is: [Given that R = 8.31 J mol−1^{-1}−1 K−1^{-1}−1]
  1. (A)350 J
  2. (B)700 J
  3. (C)748 J
  4. (D)374 J

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A ball is thrown upward with an initial velocity V0_{0}0​ from the surface of the earth. The motion of the ball is affected by a drag force equal to mγ\gammaγv2^{2}2 (where m is mass of the ball, v is its Instantneous velocity and γ\gammaγ is a constant). Time taken by the ball to rise to its zenith is:
  1. (A)1γgln⁡(1+γgV0)\dfrac{1}{\sqrt{\gamma g}}\ln\left(1+\sqrt{\dfrac{\gamma}{g}}V_{0}\right)γg​1​ln(1+gγ​​V0​)
  2. (B)1γgtan⁡−1(γgV0)\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}V_{0}\right)γg​1​tan−1(gγ​​V0​)
  3. (C)1γgsin⁡−1(γgV0)\dfrac{1}{\sqrt{\gamma g}}\sin^{-1}\left(\sqrt{\dfrac{\gamma}{g}}V_{0}\right)γg​1​sin−1(gγ​​V0​)
  4. (D)12γgtan⁡−1(2γgV0)\dfrac{1}{\sqrt{2\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{2\gamma}{g}}V_{0}\right)2γg​1​tan−1(g2γ​​V0​)

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
A thin disc of mass M and radius R has mass per unit area σ(r)=kr2\sigma(r) = kr^{2}σ(r)=kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:
  1. (A)MR22\dfrac{MR^{2}}{2}2MR2​
  2. (B)MR23\dfrac{MR^{2}}{3}3MR2​
  3. (C)MR26\dfrac{MR^{2}}{6}6MR2​
  4. (D)2MR23\dfrac{2MR^{2}}{3}32MR2​

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
Two coaxial discs, having moments of inertia I1_{1}1​ and I12\dfrac{I_{1}}{2}2I1​​, area rotating with respectively angular velocities ω1\omega_{1}ω1​ and ω12\dfrac{\omega_{1}}{2}2ω1​​, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef_{f}f​ and Ei_{i}i​ are the final and initial total energies, then (Ef_{f}f​ - Ei_{i}i​) is:
  1. (A)I1ω126\dfrac{I_{1}\omega_{1}^{2}}{6}6I1​ω12​​
  2. (B)38I1ω12\dfrac{3}{8}I_{1}\omega_{1}^{2}83​I1​ω12​
  3. (C)−I1ω1212-\dfrac{I_{1}\omega_{1}^{2}}{12}−12I1​ω12​​
  4. (D)−I1ω1224-\dfrac{I_{1}\omega_{1}^{2}}{24}−24I1​ω12​​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1_{1}1​, while water rises by the same amount h in a capillary tube of radius r2_{2}2​. The ratio, (r1_{1}1​/r2_{2}2​), is then close to:
  1. (A)3/5
  2. (B)4/5
  3. (C)2/3
  4. (D)2/5

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of to given capacitors. The capacitances are:
  1. (A)40 µF and 10 µF
  2. (B)50 µF and 30 µF
  3. (C)60 µF and 40 µF
  4. (D)20 µF and 30 µF

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let rp_{p}p​, re_{e}e​ and rHe_{He}He​ be their respective radii, then,
  1. (A)re_{e}e​ > rp_{p}p​ = rHe_{He}He​
  2. (B)re_{e}e​ > rp_{p}p​ > rHe_{He}He​
  3. (C)re_{e}e​ < rp_{p}p​ < rHe_{He}He​
  4. (D)re_{e}e​ < rp_{p}p​ = rHe_{He}He​

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
The value of acceleration due to gravity at Earth's surface is 9.8 ms−2^{-2}−2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms−2^{-2}−2, is close to: (Radius of earth = 6.4 ×\times× 106^{6}6 m)
  1. (A)6.4 ×\times× 106^{6}6 m
  2. (B)9.0 ×\times× 106^{6}6 m
  3. (C)2.6 ×\times× 106^{6}6 m
  4. (D)1.6 ×\times× 106^{6}6 m

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
In the given circuit, an ideal voltmeter connected across the 10 Ω resistance reads 2 V. The internal resistance r, of each cell is:
  1. (A)1 Ω
  2. (B)0.5 Ω
  3. (C)1.5 Ω
  4. (D)0 Ω

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
The electric field of a plane electromagnetic wave is given by E⃗=E0i^cos⁡(kz)cos⁡(ωt)\vec{E} = E_{0}\hat{i}\cos(kz)\cos(\omega t)E=E0​i^cos(kz)cos(ωt) The corresponding magnetic field B⃗\vec{B}B is then given by:
  1. (A)B⃗=E0Cj^sin⁡(kz)cos⁡(ωt)\vec{B} = \dfrac{E_{0}}{C}\hat{j}\sin(kz)\cos(\omega t)B=CE0​​j^​sin(kz)cos(ωt)
  2. (B)B⃗=E0Ck^sin⁡(kz)cos⁡(ωt)\vec{B} = \dfrac{E_{0}}{C}\hat{k}\sin(kz)\cos(\omega t)B=CE0​​k^sin(kz)cos(ωt)
  3. (C)B⃗=E0Cj^cos⁡(kz)sin⁡(ωt)\vec{B} = \dfrac{E_{0}}{C}\hat{j}\cos(kz)\sin(\omega t)B=CE0​​j^​cos(kz)sin(ωt)
  4. (D)B⃗=E0Cj^sin⁡(kz)sin⁡(ωt)\vec{B} = \dfrac{E_{0}}{C}\hat{j}\sin(kz)\sin(\omega t)B=CE0​​j^​sin(kz)sin(ωt)

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance Is 100 Ω\OmegaΩ and the output load resistance is 10 kΩ\OmegaΩ. the common emitter current gain β\betaβ is:
  1. (A)6×1026 \times 10^{2}6×102
  2. (B)10210^{2}102
  3. (C)60
  4. (D)10410^{4}104

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure. Which of the readings is inconsistent?
Sl. No.R (Ω\OmegaΩ)l (cm)
1.100060
2.10013
3.101.5
4.11.0
  1. (A)4
  2. (B)3
  3. (C)2
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coils is 10 A, then the input voltage and current in the primary coil are:
  1. (A)440 V and 5 A
  2. (B)440 and 20 A
  3. (C)220 V and 20 A
  4. (D)220 V and 10 A

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centered at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is:
  1. (A)2m(15q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{1}{5}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(51​4πϵ0​aq2​)1/2​
  2. (B)2m(115q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{1}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(151​4πϵ0​aq2​)1/2​
  3. (C)2m(415q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{4}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(154​4πϵ0​aq2​)1/2​
  4. (D)2m(215q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{2}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(152​4πϵ0​aq2​)1/2​

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
A current of 5 A passes through a copper conductor (resistivity = 1.7×10−81.7 \times 10^{-8}1.7×10−8 Ω\OmegaΩm) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1×10−31.1 \times 10^{-3}1.1×10−3 m/s.
  1. (A)1.8 m2^{2}2/Vs
  2. (B)1.0 m2^{2}2/Vs
  3. (C)1.3 m2^{2}2/Vs
  4. (D)1.5 m2^{2}2/Vs

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
n-moles of an ideal gas with constant volume heat capacity CVC_{V}CV​ undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is
  1. (A)nRCV−nR\dfrac{nR}{C_{V} - nR}CV​−nRnR​
  2. (B)nRCV+nR\dfrac{nR}{C_{V} + nR}CV​+nRnR​
  3. (C)4nRCV+nR\dfrac{4nR}{C_{V} + nR}CV​+nR4nR​
  4. (D)4nRCV−nR\dfrac{4nR}{C_{V} - nR}CV​−nR4nR​

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are:
  1. (A)4 ; 2×1082 \times 10^{8}2×108 Hz
  2. (B)4 ; 1×1081 \times 10^{8}1×108 Hz
  3. (C)0.25 ; 1×1081 \times 10^{8}1×108 Hz
  4. (D)0.25 ; 2×1082 \times 10^{8}2×108 Hz

Correct answer: (D)

Step-by-step solution →
Q21·PhysicsSingle correct
A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is :
  1. (A)2a+2b32a + \dfrac{2b}{\sqrt{3}}2a+3​2b​
  2. (B)2a+2b32a + \dfrac{2b}{3}2a+32b​
  3. (C)23a+2b\dfrac{2\sqrt{3}}{a} + 2ba23​​+2b
  4. (D)2a+2b2a + 2b2a+2b

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correct
The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos⁡(10πt+φ)x(t) = e^{-0.1t}\cos(10\pi t + \varphi)x(t)=e−0.1tcos(10πt+φ). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
  1. (A)13 s
  2. (B)27 S
  3. (C)4 s
  4. (D)7 s

Correct answer: (D)

Step-by-step solution →
Q23·PhysicsSingle correct
Two radioactive materials A and B have decay constants 10λ10\lambda10λ and λ\lambdaλ, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of a to that of B will be 1/e after a time:
  1. (A)111λ\dfrac{1}{11\lambda}11λ1​
  2. (B)110λ\dfrac{1}{10\lambda}10λ1​
  3. (C)19λ\dfrac{1}{9\lambda}9λ1​
  4. (D)1110λ\dfrac{11}{10\lambda}10λ11​

Correct answer: (C)

Step-by-step solution →
Q24·PhysicsSingle correct
One plano-convex and one plano-concave lens of same radius of curvature 'R' but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is μ1\mu_{1}μ1​ and that of 2 is μ2\mu_{2}μ2​, then the focal length of the combination is
  1. (A)R2(μ1−μ2)\dfrac{R}{2(\mu_{1} - \mu_{2})}2(μ1​−μ2​)R​
  2. (B)2Rμ1−μ2\dfrac{2R}{\mu_{1} - \mu_{2}}μ1​−μ2​2R​
  3. (C)Rμ1−μ2\dfrac{R}{\mu_{1} - \mu_{2}}μ1​−μ2​R​
  4. (D)R2−(μ1−μ2)\dfrac{R}{2 - (\mu_{1} - \mu_{2})}2−(μ1​−μ2​)R​

Correct answer: (C)

Step-by-step solution →

Chemistry — JEE Main 10 April 2019 Shift 1

Q25·ChemistrySingle correct
The species that can have a trans-isomer is: (en = ethane-1,2-diamine, ox = oxalate)
  1. (A)[Pt(en)Cl2_22​]
  2. (B)[Pt(en)2_22​Cl2_22​]2+^{2+}2+
  3. (C)[Cr(en)2_22​(ox)]+^{+}+
  4. (D)[Zn(en)Cl2_22​]

Correct answer: (B)

Step-by-step solution →
Q26·ChemistrySingle correct
The major product of the following reaction is CH3_33​CH(OH)CH2_22​CH2_22​NH2_22​ →ethyl formate (1 equiv.), triethylamine\xrightarrow{\text{ethyl formate (1 equiv.), triethylamine}}ethyl formate (1 equiv.), triethylamine​
  1. (A)CH3_33​CH(OH)CH=CH2_22​
  2. (B)HCO-O-CH(CH3_33​)CH2_22​CH2_22​NH2_22​
  3. (C)CH3_33​CH=CH-CH2_22​NH2_22​
  4. (D)CH3_33​CH(OH)CH2_22​CH2_22​NHCHO

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
The synonym for water gas when used in the production of methanol is
  1. (A)syn gas
  2. (B)natural gas
  3. (C)fuel gas
  4. (D)laughing gas

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
The regions of the atmosphere, where clouds form and where we live, respectively, are:
  1. (A)Troposphere and Stratosphere
  2. (B)Troposphere and Troposphere
  3. (C)Stratosphere and Stratosphere
  4. (D)Stratosphere and Troposphere

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correct
Consider the hydrated ions of Ti2+^{2+}2+, V2+^{2+}2+, Ti3+^{3+}3+ and Sc3+^{3+}3+. The correct order of their spin-only magnetic moments is:
  1. (A)Ti3+^{3+}3+ < T2+^{2+}2+ < Sc3+^{3+}3+ < V2+^{2+}2+
  2. (B)Sc3+^{3+}3+ < Ti3+^{3+}3+ < Ti2^{2}2 < V2+^{2+}2+
  3. (C)V2+^{2+}2+ < Ti2+^{2+}2+ < Ti3+^{3+}3+ < Sc3+^{3+}3+
  4. (D)Sc3+^{3+}3+ < Ti3+^{3+}3+ < V2+^{2+}2+ < Ti2+^{2+}2+

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correct
Amylopectin is composed of
  1. (A)α\alphaα-D-glucose, C1_11​-C4_44​ and C2_22​-C6_66​ linkages
  2. (B)β\betaβ-D-glucose, C1_11​-C4_44​ and C2_22​-C6_66​ linkages
  3. (C)α\alphaα-D-glucose, C1_11​-C4_44​ and C1_11​-C6_66​ linkages
  4. (D)β\betaβ-D-glucose, C1_11​-C4_44​ and C1_11​-C6_66​ linkages

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation xm=kp0.5\frac{x}{m} = kp^{0.5}mx​=kp0.5 Adsorption of the gas increases with:
  1. (A)Increase in p and decrease in T
  2. (B)Increase in p and increase in T
  3. (C)Decrease in p and increase in T
  4. (D)Decrease in p and decrease in T

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
A bacterial infection in an internal would grows as N'(t) = N0_00​ exp(t), where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as dNdt=−5N2\frac{dN}{dt} = -5N^2dtdN​=−5N2. What will be the plot of N0N\frac{N_0}{N}NN0​​ vs, t after 1 hour?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
The graph between ∣ψ∣2|\psi|^{2}∣ψ∣2 and r(radial distance) is shown below. This represents:
  1. (A)1s orbital
  2. (B)3s orbital
  3. (C)2s orbital
  4. (D)2p orbital

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Consider the following table: a and b are van der Waals constants. The correct statement about the gases is:
Gasa / (k Pa dm6^66 mol−1^{-1}−1)b / (dm3^33 mol−1^{-1}−1)
A642.320.05196
B155.210.04136
C431.910.05196
D155.210.4382
  1. (A)Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D
  2. (B)Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D.
  3. (C)Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D
  4. (D)Gas C will occupy more volume than gas A; gas B will be more compressible than gas D.

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
The principle of column chromatography is
  1. (A)Differential absorption of the substances on the solid phase.
  2. (B)Differential adsorption of the substances on the solid phase.
  3. (C)Gravitational force.
  4. (D)Capillary action.

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2_22​ for complete combustion, and 40 mL of CO2_22​ is formed. The formula of the hydrocarbon is:
  1. (A)C4_44​H10_{10}10​
  2. (B)C4_44​H6_66​
  3. (C)C4_44​H7_77​Cl
  4. (D)C4_44​H8_88​

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
Three complexes, [CoCl(NH3_33​)5_55​]2+^{2+}2+ (I), [co(NH3_33​)5_55​H2_22​O]3+^{3+}3+ (II) and [Co(NH3_33​)6_66​]3+^{3+}3+ (III) Absorb light in the visible region. The correct order of the wavelength of light absorbed by them is
  1. (A)(II) > (I) > (III)
  2. (B)(III) > (II) > (I)
  3. (C)(I) > (II) > (III)
  4. (D)(III) > (I) > (II)

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Major products of the following reaction are: + HCHO →(i) 50% NaOH, (ii) H3O+\xrightarrow{\text{(i) 50\% NaOH, (ii) H}_3\text{O}^{+}}(i) 50% NaOH, (ii) H3​O+​
  1. (A)CH3_33​OH and HCO2_22​H
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
The major product of the following reaction is: →ΔHI (excess)\xrightarrow[\Delta]{\text{HI (excess)}}HI (excess)Δ​
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Match the refining methods (Column I) with metals (Column II).
Column I (Refining methods)Column II (Metals)
I.Liquationa.Zr
II.zone Refiningb.Ni
III.Mond processC.Sn
IV.Van Arkel Methodd.Ga
  1. (A)(I) \u2013 (b); (II) \u2013 (c); (III) \u2013 (d); (IV) \u2013 (a)
  2. (B)(I) \u2013 (b); (II) \u2013 (d); (III) \u2013 (a); (IV) \u2013 (c)
  3. (C)(I) \u2013 (c); (II) \u2013 (a); (III) \u2013 (b); (IV) \u2013 (d)
  4. (D)(I) \u2013 (c); (II) \u2013 (d); (III) \u2013 (b); (IV) \u2013 (a)

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
The increasing order of the reactivity of the following compounds towards electrophilic aromatic substitution reaction is:
  1. (A)(III) < (I) < (II)
  2. (B)(III) < (II) < (I)
  3. (C)(II) < (I) < (III)
  4. (D)(I) < (III) < (II)

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
A process will be spontaneous at all temperatures if:
  1. (A)ΔH < 0 and ΔS < 0
  2. (B)ΔH < 0 and ΔS > 0
  3. (C)ΔH > 0 and ΔS > 0
  4. (D)ΔH > 0 and ΔS < 0

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
Consider the following statements (a) The pH of a mixture containing 400 mL of 0.1 M H2SO4H_2SO_4H2​SO4​ and 400 mL of 0.1 M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with Ka=10−5K_a = 10^{-5}Ka​=10−5 has a pH = 5. the degree of dissociation of this acid is 50%. (d) The Le Chatelier's principle is not applicable to common-ion effect. The correct statements are:
  1. (A)(a) and (b)
  2. (B)(a), (b) and (c)
  3. (C)(b) and (c)
  4. (D)(a), (b) and (d)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
Ethylamine (C2H5NH2C_2H_5NH_2C2​H5​NH2​) can be obtained from N-ethylphthalimide on treatment with:
  1. (A)CaH2CaH_2CaH2​
  2. (B)H2OH_2OH2​O
  3. (C)NaBH4NaBH_4NaBH4​
  4. (D)NH2NH2NH_2NH_2NH2​NH2​

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correct
The correct order of catenation is:
  1. (A)C > Sn > Si ≈ Ge
  2. (B)Ge > Sn > Si > C
  3. (C)Si > Sn > C > Ge
  4. (D)C > Si > Ge ≈ Sn

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
The major product of the following reaction is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
Which of the following is a condensation polymer?
  1. (A)Nylon 6, 6
  2. (B)Neoprene
  3. (C)Buna–S
  4. (D)Teflon

Correct answer: (A)

Step-by-step solution →
Q48·ChemistrySingle correct
Consider the statements S1 and S2: S1: Conductivity always increases with decrease in the concentration of electrolyte. S2: Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is:
  1. (A)S1 is wrong and S2 is correct
  2. (B)Both S1 and S2 are wrong
  3. (C)S1 is correct and S2 is wrong
  4. (D)Both S1 and S2 are correct

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
Increasing rate of SN_NN​1 reaction in the following compounds is
  1. (A)(A) < (B) < (C) < (D)
  2. (B)(A) < (B) < (D) < (C)
  3. (C)(B) < (A) < (D) < (C)
  4. (D)(B) < (A) < (C) < (D)

Correct answer: (D)

Step-by-step solution →
Q50·ChemistrySingle correct
The oxoacid of sulphur that does not contain bond between sulphur atoms is:
  1. (A)H2S2O3H_2S_2O_3H2​S2​O3​
  2. (B)H2S2O4H_2S_2O_4H2​S2​O4​
  3. (C)H2S2O7H_2S_2O_7H2​S2​O7​
  4. (D)H2S4O6H_2S_4O_6H2​S4​O6​

Correct answer: (C)

Step-by-step solution →
Q51·ChemistrySingle correct
The isoelectronic set of ions is
  1. (A)Li+^++, Na+^++, O2−^{2-}2− and F−^-−
  2. (B)F−^-−, Li+^++, Na+^++ and Mg2+^{2+}2+
  3. (C)N3−^{3-}3−, O2−^{2-}2−, F−^-− and Na+^++
  4. (D)N3−^{3-}3−, Li+^++, Mg2+^{2+}2+ and O2−^{2-}2−

Correct answer: (C)

Step-by-step solution →
Q52·ChemistrySingle correct
The alloy used in the construction of aircrafts is
  1. (A)Mg–Mn
  2. (B)Mg-Zn
  3. (C)Mg-Al
  4. (D)Mg-Sn

Correct answer: (C)

Step-by-step solution →
Q53·ChemistrySingle correct
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea = 60 g mol−1^{-1}−1)
  1. (A)0.027 mmHg
  2. (B)0.031 mmHg
  3. (C)0.028 mmHg
  4. (D)0.017 mmHg

Correct answer: (D)

Step-by-step solution →

Mathematics — JEE Main 10 April 2019 Shift 1

Q54·MathematicsSingle correct
If the coefficients of x2^{2}2 and x3^{3}3 are both zero, in the expansion of the expression (1 + ax + bx2^{2}2) (1 − 3x)15^{15}15 in powers of x, then the ordered pair (a, b) is equal to
  1. (A)(−54, 315)
  2. (B)(28, 861)
  3. (C)(28, 315)
  4. (D)(−21, 714)

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correct
The sum 3×112+5×(13+23)12+22+7×(13+23+33)12+22+32+........\frac{3 \times 1}{1^{2}} + \frac{5 \times (1^{3}+2^{3})}{1^{2}+2^{2}} + \frac{7 \times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}} + ........123×1​+12+225×(13+23)​+12+22+327×(13+23+33)​+........ upto 10th^{th}th term, is
  1. (A)620
  2. (B)660
  3. (C)680
  4. (D)600

Correct answer: (B)

Step-by-step solution →
Q56·MathematicsSingle correct
Which one of the following Boolean expressions is a tautology?
  1. (A)(p ∨ q) ∧ (p∨∼q)
  2. (B)(p ∧ q) ∨ (p∧∼q)
  3. (C)(p ∨ q) ∧ (∼p∨∼q)
  4. (D)(p ∨ q) ∨ (p∨∼q)

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
If α and β are the roots of the quadratic equation, x2^{2}2 + x sinθ − 2sinθ = 0, θ ∈ (0, π2\frac{\pi}{2}2π​) then α12+β12(α−12+β−12)(α−β)24\frac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha-\beta)^{24}}(α−12+β−12)(α−β)24α12+β12​ is equal to:
  1. (A)212(sin⁡θ+8)12\frac{2^{12}}{(\sin\theta+8)^{12}}(sinθ+8)12212​
  2. (B)212(sin⁡θ−4)12\frac{2^{12}}{(\sin\theta-4)^{12}}(sinθ−4)12212​
  3. (C)212(sin⁡θ−8)6\frac{2^{12}}{(\sin\theta-8)^{6}}(sinθ−8)6212​
  4. (D)26(sin⁡θ+8)12\frac{2^{6}}{(\sin\theta+8)^{12}}(sinθ+8)1226​

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correct
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot−1^{-1}−1(32\sqrt{2}2​) and cosec−1^{-1}−1(22\sqrt{2}2​) respectively, then the height of the tower (in metres) is:
  1. (A)25
  2. (B)10510\sqrt{5}105​
  3. (C)10033\frac{100}{3\sqrt{3}}33​100​
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q59·MathematicsSingle correct
If a > 0 and z = (1+i)2a−i\frac{(1+i)^{2}}{a-i}a−i(1+i)2​, has magnitude 25\sqrt{\frac{2}{5}}52​​, then zˉ\bar{z}zˉ is equal to:
  1. (A)−35−15i-\frac{3}{5}-\frac{1}{5}i−53​−51​i
  2. (B)−15−35i-\frac{1}{5}-\frac{3}{5}i−51​−53​i
  3. (C)−15+35i-\frac{1}{5}+\frac{3}{5}i−51​+53​i
  4. (D)15−35i\frac{1}{5}-\frac{3}{5}i51​−53​i

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
If y = 1f(x) is the solution of the differential equation dydx=(tan⁡x−y)sec⁡2x\frac{dy}{dx}=(\tan x - y)\sec^{2}xdxdy​=(tanx−y)sec2x, x ∈ (−π2,π2-\frac{\pi}{2}, \frac{\pi}{2}−2π​,2π​), such that y(0) = 0, then y(−π4-\frac{\pi}{4}−4π​) is equal to
  1. (A)12−e\frac{1}{2}-e21​−e
  2. (B)1e−2\frac{1}{e}-2e1​−2
  3. (C)e−2e-2e−2
  4. (D)2+1e2+\frac{1}{e}2+e1​

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let f: R → R be differentiable at c ∈ R and f(c) = 0. If g(x) = |f(x)|, then at x = c, g is:
  1. (A)differentiable if f′(c) = 0
  2. (B)differentiable if f′(c) ≠ 0
  3. (C)not differentiable
  4. (D)not differentiable if f′(c) = 0

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
Let f(x) = x2^{2}2, x ∈ R. For any A ⊆ R, define g(A) = {x ∈ R : f(x) ∈ A}. If S = [0, 4], then which one of the following statements is not true?
  1. (A)f(g(S)) ≠ f(S)
  2. (B)f(g(S)) = S
  3. (C)g(f(S)) ≠ S
  4. (D)g(f(S)) = g(S)

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
The number of 6 digit numbers hat can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digits is repeated, is
  1. (A)36
  2. (B)60
  3. (C)72
  4. (D)48

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, −3), then its radius is
  1. (A)323\sqrt{2}32​
  2. (B)3
  3. (C)2
  4. (D)222\sqrt{2}22​

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correct
Let f(x) = ex−xe^{x}-xex−x and g(x) = x2−xx^{2}-xx2−x, ∀\forall∀ x ∈\in∈ R. Then the set of all x ∈\in∈ R, where the function h(x) = (fog) (x) is increasing is:
  1. (A)[0,12]∪[1,∞)\left[0,\dfrac{1}{2}\right] \cup [1,\infty)[0,21​]∪[1,∞)
  2. (B)[1,12]∪[12,∞)\left[1,\dfrac{1}{2}\right] \cup \left[\dfrac{1}{2},\infty\right)[1,21​]∪[21​,∞)
  3. (C)[−12,0]∪[1,∞)\left[\dfrac{-1}{2},0\right] \cup [1,\infty)[2−1​,0]∪[1,∞)
  4. (D)[0,∞)[0,\infty)[0,∞)

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
The value of ∫02π[sin⁡2x(1+cos⁡3x)] dx\displaystyle\int_{0}^{2\pi} [\sin 2x(1+\cos 3x)]\, dx∫02π​[sin2x(1+cos3x)]dx where [t] denotes the greatest integer function, is:
  1. (A)π\piπ
  2. (B)−2π-2\pi−2π
  3. (C)2π2\pi2π
  4. (D)−π-\pi−π

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
If the circles x2+y2+5Kx+2y+K=0x^{2}+y^{2}+5Kx+2y+K=0x2+y2+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y−1=02(x^{2}+y^{2})+2Kx+3y-1=02(x2+y2)+2Kx+3y−1=0, (K∈\in∈R), intersect at the points P and Q, then the line 4x + 5y − K = 0 passes through P and Q for:
  1. (A)exactly one value of K
  2. (B)not value of K
  3. (C)infinitely many values of K
  4. (D)exactly two values of K

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
If a1_{1}1​, a2_{2}2​, a3_{3}3​ ........... an_{n}n​ are in A.P and a1_{1}1​ + a4_{4}4​ + a7_{7}7​ + ............... + a16_{16}16​ = 114, then a1_{1}1​ + a6_{6}6​ + a11_{11}11​ + a16_{16}16​ is equal to
  1. (A)76
  2. (B)64
  3. (C)98
  4. (D)38

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
If for some x ∈\in∈ R, the frequency distribution of the marks obtained by 20 students in a test is Then the mean of the marks is:
Marks2357
Frequency(x+1)2(x+1)^{2}(x+1)22x − 5x2−3xx^{2}-3xx2−3xx
  1. (A)2.8
  2. (B)3.2
  3. (C)2.5
  4. (D)3.0

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\displaystyle\lim_{x\to 1}\dfrac{x^{4}-1}{x-1}=\lim_{x\to k}\dfrac{x^{3}-k^{3}}{x^{2}-k^{2}}x→1lim​x−1x4−1​=x→klim​x2−k2x3−k3​, then k is
  1. (A)38\dfrac{3}{8}83​
  2. (B)83\dfrac{8}{3}38​
  3. (C)43\dfrac{4}{3}34​
  4. (D)32\dfrac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
All the pairs (x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5⋅14sin⁡2y≤12^{\sqrt{\sin^{2}x-2\sin x+5}}\cdot\dfrac{1}{4^{\sin^{2}y}}\le 12sin2x−2sinx+5​⋅4sin2y1​≤1 also Satisfy the equation:
  1. (A)2∣sin⁡x∣=3sin⁡y2|\sin x|=3\sin y2∣sinx∣=3siny
  2. (B)sin⁡x=∣sin⁡y∣\sin x=|\sin y|sinx=∣siny∣
  3. (C)2sin⁡x=sin⁡y2\sin x=\sin y2sinx=siny
  4. (D)sin⁡x=2sin⁡y\sin x=2\sin ysinx=2siny

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
lim⁡n→∞((n+1)1/3n4/3+(n+2)1/3n4/3+....+(2n)1/3n4/3)\displaystyle\lim_{n\to\infty}\left(\dfrac{(n+1)^{1/3}}{n^{4/3}}+\dfrac{(n+2)^{1/3}}{n^{4/3}}+....+\dfrac{(2n)^{1/3}}{n^{4/3}}\right)n→∞lim​(n4/3(n+1)1/3​+n4/3(n+2)1/3​+....+n4/3(2n)1/3​) is equal to:
  1. (A)34(2)4/3−34\dfrac{3}{4}(2)^{4/3}-\dfrac{3}{4}43​(2)4/3−43​
  2. (B)43(2)3/4\dfrac{4}{3}(2)^{3/4}34​(2)3/4
  3. (C)34(2)4/3−43\dfrac{3}{4}(2)^{4/3}-\dfrac{4}{3}43​(2)4/3−34​
  4. (D)43(2)4/3\dfrac{4}{3}(2)^{4/3}34​(2)4/3

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
If a directrix of a hyperbola centered at the origin and passing through the point (4,−23)(4,-2\sqrt{3})(4,−23​) is 5x=455x=4\sqrt{5}5x=45​ and its eccentricity is e, then
  1. (A)4e4+8e2−35=04e^{4}+8e^{2}-35=04e4+8e2−35=0
  2. (B)4e4−24e2+35=04e^{4}-24e^{2}+35=04e4−24e2+35=0
  3. (C)4e4−12e2−27=04e^{4}-12e^{2}-27=04e4−12e2−27=0
  4. (D)4e4−24e2+27=04e^{4}-24e^{2}+27=04e4−24e2+27=0

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correct
The region represented by ∣x−y∣≤2|x-y|\le 2∣x−y∣≤2 and ∣x+y∣≤2|x+y|\le 2∣x+y∣≤2 is bounded by a:
  1. (A)rhombus of area 828\sqrt{2}82​ sq. units
  2. (B)square of area 16 sq. units
  3. (C)rhombus of side length 2 units
  4. (D)square of side length 222\sqrt{2}22​ units

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is:
  1. (A)110\dfrac{1}{10}101​
  2. (B)117\dfrac{1}{17}171​
  3. (C)112\dfrac{1}{12}121​
  4. (D)111\dfrac{1}{11}111​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
Let A(3, 0, −1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos(∠\angle∠GOA) (O being he origin) is equal to:
  1. (A)130\dfrac{1}{\sqrt{30}}30​1​
  2. (B)1215\dfrac{1}{2\sqrt{15}}215​1​
  3. (C)1610\dfrac{1}{6\sqrt{10}}610​1​
  4. (D)115\dfrac{1}{\sqrt{15}}15​1​

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
If ∫dx(x2−2x+10)2=A(tan⁡−1(x−13)+f(x)x2−2x+10)+C\displaystyle\int \dfrac{dx}{(x^{2}-2x+10)^{2}}=A\left(\tan^{-1}\left(\dfrac{x-1}{3}\right)+\dfrac{f(x)}{x^{2}-2x+10}\right)+C∫(x2−2x+10)2dx​=A(tan−1(3x−1​)+x2−2x+10f(x)​)+C where C is a constant of integration, then:
  1. (A)A=127A=\dfrac{1}{27}A=271​ and f(x)=−(x−1)f(x)=-(x-1)f(x)=−(x−1)
  2. (B)A=154A=\dfrac{1}{54}A=541​ and f(x)=9(x−1)2f(x)=9(x-1)^{2}f(x)=9(x−1)2
  3. (C)A=154A=\dfrac{1}{54}A=541​ and f(x)=3(x−1)f(x)=3(x-1)f(x)=3(x−1)
  4. (D)A=181A=\dfrac{1}{81}A=811​ and f(x)=3(x−1)f(x)=3(x-1)f(x)=3(x−1)

Correct answer: (C)

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
  • 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
  • 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
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Polymers 64/186
  • States of Matter: Gases and Liquids 52/186
  • Mathematical Reasoning 26/186
  • Communication Systems 11/186
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