JEE MAIN Question Paper with Answers & Solutions for

JEE (MAIN)-2019 (Online) Phase-2

(Physics, Chemistry and Mathematics)

1. A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρ(r)=Kr2ρ(r)=r2K? . Identify the correct relation between the radius RR of the particle's orbit and its period TT : (1) T/RT/R is a constant (2) TRTR is a constant (3) T/R2T/R2 is a constant (4) T2/R3T2/R3 is a constant

Answer (1)

Sol.M(in)=∫0R4πr2dr⋅kr2Sol.M(in)?=∫0R?4πr2dr⋅r2k? M(in)=4πkRM(in)?=4πkR G⋅4πkRR2=V2R⇒v=4πGKG⋅R24πkR?=RV2?⇒v=4πGK? T=2πRv=2π⋅R4πGKT=v2πR?=4πGK?2π⋅R? T=TR=constantT=RT?=constant

1. The specific heats, CpCp? and CvCv? of a gas of diatomic molecules, A, are given (in units of Jmol−1K−1Jmol−1K−1 ) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then:

(1) A is rigid but B has a vibrational mode.
(2) A has a vibrational mode but B has none.
(3) Both A and B have a vibrational mode each.
(4) A has one vibrational mode and B has two.

Answer (2)

Sol. For (A) Cp=29Cp?=29 Cv=22Cv?=22

For(B)Cp=30Cv=21For(B)Cp?=30Cv?=21∴γA=CpACvA=2922=1.31∴γA?=CvA??CpA???=2229?=1.31

When A has vibrational degree of freedom, then γA=9/7=1.29γA?=9/7=1.29

γB=CpBCvB=3021=1.42γB?=CvB??CpB???=2130?=1.42

⇒⇒ B has no vibrational degree of freedom

1. The position vector of a particle changes with time according to the relation r?(t)=15t2i^+(4−20t2)j^r(t)=15t2i^+(4−20t2)j^? . What is the magnitude of the acceleration at t=1t=1 ?

(1) 50
(2) 100

(3) 40
(4) 25

Answer (1)

Sol.r?=15t2i^+4j^−20t2j^Sol.r=15t2i^+4j^?−20t2j^? dr?dt=30ti^−40tj^dtdr?=30ti^−40tj^? d2r?dt2=30i^−40j^dt2d2r?=30i^−40j^? ∴d2r?dt2=50m/s2∴dt2d2r?=50m/s2

1. 50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%)(25%) is reflected from the surface and the rest is absorbed. The force exerted on 1m21m2 surface area will be close to c=3×108m/sc=3×108m/s :

(1) 20×10−8N20×10−8N (2) 35×10−8N35×10−8N (3) 15×10−8N15×10−8N (4) 10×10−8N10×10−8N

Answer (1)

Sol. P=WcP=cW? and pressure =1/c=1/c ∴hλ+h4λ⇒ΔP=5h4λ∴λh?+4λh?⇒ΔP=4λ5h? for one photon ∴54NhλΔtA=∴45?λΔtANh?= pressure But (Nhcλ⋅ΔtA)=50W/m2(λ⋅ΔtANhc?)=50W/m2 ∴54×50c=∴45?×c50?= pressure ∴Fn=5×50×1m24×c=20×10−8N∴Fn?=4×c5×50×1m2?=20×10−8N

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1. A metal wire of resistance 3Ω3Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60?60? at the centre, the equivalent resistance between these two points will be:

53Ω72Ω(2)52Ω(4)125Ω(4)125Ω(4) Ω (4) Ω (4) Ω 35?Ω27?Ω?(2)25?Ω(4)512?Ω(4)512?Ω(4) Ω (4) Ω (4) Ω 

Answer (1)

Sol.

[image]
.

R0=ρlA=3ΩR0?=ρAl?=3Ω

Now if l=2ll=2l

Then A=A2A=2A?

∴R=ρ2l×2A=∴R=Aρ2l×2?=R1=126=2ΩR1?=612?=2ΩR2=10ΩR2?=10Ω∴1Req=12+110∴Req?1?=21?+101?∴Req=53Ω∴Req?=35?Ω

1. A thin smooth rod of length L and mass M is rotating freely with angular speed ω0ω0? about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be:

 (1)  M ω0 M +3 m  (2)  M ω0 M + m  (3)  M ω0 M +6 m  (4)  M ω0 M +2 m (4) (1)  M +3 m  M ω0?? (2)  M + m  M ω0?? (3)  M +6 m  M ω0?? (4)  M +2 m  M ω0???(4)

Answer (3)

Sol. Initial angular momentum == Final Angular Momentum

ML212ω0=(ML212+2ML24)ω12ML2?ω0?=(12ML2?+24ML2?)ω ⇒ω=Mω0M+6m⇒ω=M+6mMω0??

1. Moment of inertia of a body about a given axis is 1.5kgm21.5kgm2 Initially the body is at rest. In order to produce a rotational kinetic energy of 1200J1200J the angular acceleration of 20rad/s220rad/s2 must be applied about the axis for a duration of

(1) 2.5s2.5s (2) 2s2s (3) 5s5s (4) 3s3s

Answer (2)

Sol. 1=1501=150 α=20rad/s2α=20rad/s2

∴ω=αt∴ω=αt  E=12ω02=1200 E=21?ω02?=1200  12×1.5×(20t)2=1200 J  21?×1.5×(20t)2=1200 J  ⇒ t=2 s ⇒ t=2 s 

1. The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0−0.5A0−0.5A ?

(1) 0.2ohm0.2ohm (2) 0.002ohm0.002ohm (3) 0.5ohm0.5ohm (4) 0.02ohm0.02ohm

Answer (1)

[image]
We have

 I g R g=(0.5− I g) S ⇒(0.002)(50)=(0.5− I g) S = S =0.002×500.5=0.2Ω(4) I g? R g?=(0.5− I g?) S ⇒(0.002)(50)=(0.5− I g?) S = S =0.50.002×50?=0.2Ω?(4)

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1. A He ++ ion is in its first excited state. Its ionization energy is :

(1) 13.60 eV
(2) 6.04 eV
(3) 48.36 eV
(4) 54.40 eV

Answer (1)

Sol.En:−E0z2n2=−E0×44=−E0Sol.En?:n2−E0?z2?=4−E0?×4?=−E0?

To ionise it E0E0? energy must be supplied.

∴E0=13.6eV.∴E0?=13.6eV.

1. Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20ms−120ms−1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B?

(speed of sound in air =340ms−1=340ms−1

(1) 2150 Hz
(2) 2300 Hz
(3) 2060 Hz
(4) 2250 Hz

Answer (4)

Sol.S20m/sO20m/sO20m/sSol.20m/sS?20m/sO?20m/sO? f=(ν±u0)(ν±us)⋅f0=(ν−20)(ν+20)⋅f0f=(ν±us?)(ν±u0?)?⋅f0?=(ν+20)(ν−20)?⋅f0? ⇒2000=320360⋅f0⇒2000=360320?⋅f0? ⇒2000×98=f0=2250Hz.⇒82000×9?=f0?=2250Hz.

1. A convex lens of focal length 20 cm20 cm produces images of the same magnification 2 when an object is kept at two distances x1x1? and x2x2? (x1>x2)(x1?>x2?) from the lens. The ratio of x1x1? and x2x2? is:

(1) 3:1
(2) 2:1
(3) 4:3
(4) 5:3

Answer (1)

Sol.∴1ν−1u=1fSol.∴ν1?−u1?=f1? ∴ν=(±2u)∴ν=(±2u) ∴12u+1u=120⇒32u=120∴2u1?+u1?=201?⇒2u3?=201? ∴u1=30cm∴u1?=30cm And1u−12u=120Andu1?−2u1?=201?

∴12u=120∴u2=10∴2u1?=201?∴u2?=10 ∴3010=3.∴1030?=3.

1. The position of a particle as a function of time t, is given by

x(t)=at+bt2−ct3x(t)=at+bt2−ct3

where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

(1)a+b24c(2)a+b23c(1)a+4cb2?(2)a+3cb2? (3)a+b22c(4)a+b2c(3)a+2cb2?(4)a+cb2?

Answer (2)

Sol.x=at+bt2−ct3Sol.x=at+bt2−ct3x?=a+2bt−3ct2x?=a+2bt−3ct2x?=2b−6ctx?=2b−6ctForx?=0t=+b3cForx?=0t=+3cb?∴ν=x?=a+2b(+b3c)−3c(b23c×3c)∴ν=x?=a+2b(3c+b?)−3c(3c×3cb2?)⇒ν=−b23c+2b23c+a=a+b23c⇒ν=−3cb2?+3c2b2?+a=a+3cb2?

1. The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :

(1)(n+1)V(K+n)(2)VK+n(1)(K+n)(n+1)V?(2)K+nV? (3)V(4)nVK+n(3)V(4)K+nnV?

Answer (1)

Sol. Initially

Q=CV(1+n)Q=CV(1+n) ∴Ceq=(K+n)C∴Ceq?=(K+n)C ∴V=CV(1+n)(K+n)C=V(1+n)(K+n)∴V=(K+n)CCV(1+n)?=(K+n)V(1+n)?

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1. The physical sizes of the transmitter and receiver antenna in a communication system are:

(1) Inversely proportional to modulation frequency
(2) Inversely proportional to carrier frequency
(3) Proportional to carrier frequency
(4) Independent of both carrier and modulation frequency

Answer (2)

Sol. Size of antenna depends on wavelength of carrier wave.

1. A wedge of mass M=4mM=4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by :

v2g2v25g2v27g2v27g(4)gv2?5g2v2?7g2v2?7g2v2??(4)

Answer (2)

Sol. mv=(4m+m)vmv=(4m+m)v

 

1. A wooden block floating in a bucket of water has 4554? of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is

(1) 0.5
(2) 0.8
(3) 0.7
(4) 0.6

Answer (4)

Volvg=45vρoggVolvg=v2ρogg+v2ρogg⇒(ρog2+ρoil2)=45ρog⇒ρoil2=ρog(45−12)=310ρog⇒ρoil=35ρog=0.6ρog(1)Volvg=54?vρog?gVolvg=2v?ρog?g+2v?ρog?g⇒(2ρog??+2ρoil??)=54?ρog?⇒2ρoil??=ρog?(54?−21?)=103?ρog?⇒ρoil?=53?ρog?=0.6ρog??(1)

1. Two materials having coefficients of thermal conductivity ?3K′?3K′ and K¨′K¨′ and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ?02′?02′? and ?01′?01′? respectively, (02>01)(02?>01?) . The temperature at the interface is:

[image]

Answer (2)

Sol. H=3KAd(02−0)=KA3d(0−01)Sol. H=d3KA?(02?−0)=3dKA?(0−01?) ⇒0=90210+0110⇒0=10902??+1001??

1. The area of a square is 5.29cm25.29cm2 . The area of 7 such squares taking into account the significant figures is :

(1)37.03cm2(2)37.0cm2(1)37.03cm2(2)37.0cm2 (3)37.030cm2(4)37cm2(3)37.030cm2(4)37cm2

Answer (2)

Sol. 5.29×7=37.0cm25.29×7=37.0cm2

Answer should be in 3 significant digits.

1. Diameter of the objective lens of a telescope is 250 cm250 cm . For light of wavelength 600 nm600 nm coming from a distant object, the limit of resolution of the telescope is close to :

(1) 1.5×10−71.5×10−7 rad
(2) 3.0×10−73.0×10−7 rad
(3) 2.0×10−72.0×10−7 rad
(4) 4.5×10−74.5×10−7 rad

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Answer (2)

Sol. θ=1.22λDθ=D1.22λ?

⇒θ=1.22×600×10−9250×100=2.92×10−7⇒θ=2501.22×600×10−9?×100=2.92×10−7⇒θ=3×10−7rad⇒θ=3×10−7rad

1. A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λxλx?' and 'λyλy?' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is

(1) λxλy?λx−λy??λx?−λy??λx?λy??
(2) λx−λyλx?−λy?

(3) λx+λyλx?+λy?
(4) λxλyλx+λyλx?+λy?λx?λy??

Answer (1)

Sol. P1=hλxP1?=λx?h? P2=hλyP2?=λy?h?

P=P1−P2=h(1λx−1λy)P=P1?−P2?=h(λx?1?−λy?1?)∴P=hλ∴P=λh?hλ=h(1λx−1λy)λh?=h(λx?1?−λy?1?)1λ=?λy−λx?λxλyλ1?=λx?λy??λy?−λx???λ=λxλy?λx−λy?λ=?λx?−λy??λx?λy??

1. A moving coil galvanometer has a coil with 175 turns and area 1 cm². It uses a torsion band of torsion constant 10−610−6 N-m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately

(1) 10−410−4
(2) 10−210−2

(3) 10−110−1
(4) 10−310−3

Answer (4)

Sol. NIAB = KQ

175×1×10−3×1×10−4×B=10−6×π180175×1×10−3×1×10−4×B=18010−6×π?⇒B=π180×10175≈9.97×10−4T.⇒B=180π?×17510?≈9.97×10−4T.⇒B=10−3T.⇒B=10−3T.

1. In a conductor, if the number of conduction electrons per unit volume is 8.5×1028m−38.5×1028m−3 and mean free time is 25 fs (femto second), it's approximate resistivity is

(me=9.1×10−31kg)(me?=9.1×10−31kg)

(1) 10−5Ωm10−5Ωm
(2) 10−6Ωm10−6Ωm

(3) 10−7Ωm10−7Ωm
(4) 10−8Ωm10−8Ωm

Answer (4)

Sol. J=nevdJ=nevd? ∴vd=eEτm∴vd?=meEτ?

∴J=ne⋅eE⋅τm=σE∴J=mne⋅eE⋅τ?=σEJ=ne2τmE=σEJ=mne2τ?E=σE∴ρ=mne2τ∴ρ=ne2τm?=9.1×10−318.5×1028×(1.6×10−19)2×25×10−15Ωm=8.5×1028×(1.6×10−19)2×25×10−159.1×10−31?Ωm=9.18.5×2.56×25×10−(59−53)=1.67×10−8Ωm=8.5×2.56×259.1?×10−(59−53)=1.67×10−8Ωm

1. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μlμl? is put between the lens and the mirror, the pin has to be moved to A', such that OA' = 27 cm, to get its inverted real image at A' itself. The value of μlμl? will be

[image]

(1) 22?
(2) 4334?

(3) 33?
(4) 3223?

Answer (2)

Sol. fL=18cmfL?=18cm

118=0.5×2R⇒R=18cm181?=0.5×R2?⇒R=18cm1f2=(μl−1)(−118)f2?1?=(μl?−1)(−181?)

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1. Two coils ∇⋅P′∇⋅P′ and ∇⋅Q′∇⋅Q′ are separated by some distance. When a current of 3 A flows through coil ∇⋅P′∇⋅P′ , a magnetic flux of 10−310−3 Wb passes through Q. No current is passed through ∇⋅Q′∇⋅Q′ . When no current passes through ∇⋅P′∇⋅P′ and a current of 2 A passes through ∇⋅Q′∇⋅Q′ , the flux through ∇⋅P′∇⋅P′ is

(1)6.67×10−3Wb(2)6.67×10−4Wb(3)3.67×10−3Wb(4)3.67×10−4Wb(3)?(1)6.67×10−3Wb(2)6.67×10−4Wb(3)3.67×10−3Wb(4)3.67×10−4Wb?(3)

Answer (2)

Sol. ?Q=Mi?Q?=Mi ⇒10−3=M(3)⇒10−3=M(3) ?P=M(2)?P?=M(2) ∴10−33=?P2∴310−3?=2?P?? ?P=203×10−4=6.67×10−4Wb.?P?=320?×10−4=6.67×10−4Wb.

1. A massless spring (k=800 N/m)(k=800 N/m) , attached with a mass (500 g)(500 g) is completely immersed in 1kg1kg of water. The spring is stretched by 2cm2cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass =400J/kgK=400J/kgK , specific heat of water =4184J/kgK=4184J/kgK

(1)10−5K(2)10−1K(3)10−3K(4)10−4K(4)(1)10−5K(3)10−3K?(2)10−1K(4)10−4K?(4)

Answer (1)

Sol.12kΔx2=E(dissipated)Sol.21?kΔx2=E(dissipated)∴12×800×(2×2100×100)=16100J∴21?×800×(100×1002×2?)=10016?J∴16100=12×400×ΔT+1×4184×ΔT∴10016?=21?×400×ΔT+1×4184×ΔT⇒16100=(200+4184)ΔT=4384ΔT⇒10016?=(200+4184)ΔT=4384ΔT∴ΔT=164384×100=3.6×10−5K∴ΔT=4384×10016?=3.6×10−5K

1. A particle of mass μ′m′μ′m′ is moving with speed ′2ν′′2ν′ and collides with a mass ′2m′′2m′ moving with speed ∇⋅v′∇⋅v′ in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass μ′m′μ′m′ , which move at angle 45?45? with respect to the original direction.

The speed of each of the moving particle will be

(1)ν/(22)(2)ν/2(1)ν/(22?)(2)ν/2? (3)2ν(4)22ν(3)2?ν(4)22?ν

Answer (4)

Sol. Initial momentum ⋅Pi=2mν+2mν=4mν⋅Pi?=2mν+2mν=4mν Let ν′ν′ be the speed of I particle

[image]

2mv′2=4mv22?mv′?=4mvv′=22vv′=22?v

1. A string 2.0m2.0m long and fixed at its ends is driven by a 240Hz240Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

(1) 320m/s320m/s 120Hz120Hz (2) 320m/s320m/s 80Hz80Hz (3) 180 m/s180 m/s 80 Hz80 Hz (4) 180 m/s180 m/s 120 Hz120 Hz

Answer (2)

Sol.λ2=23Sol.2λ?=32? ⇒λ=43m⇒λ=34?m ∴ν=43×240=320m/s∴ν=34?×240=320m/s 3rdharmonic3rdharmonic fn=nf0fn?=nf0? f0=2403=80Hzf0?=3240?=80Hz

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1. The logic gate equivalent to the given logic circuit is

[image]

(1) OR
(2) NAND
(3) AND
(4) NOR

Answer (1)

Sol. y=A‾⋅B‾=A+By=A⋅B=A+B

y=A+By=A+B

⇒⇒ OR gate

1. A very long solenoid of radius R is carrying current l(t)=ke−ctl(t)=ke−ct (k>0)(k>0) , as a function of time (t≥0)(t≥0) . Counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:

[image]

Answer (2)

Sol. i=te−ct⋅ki=te−ct⋅k

\begin{array}{rl} & {\therefore \phi = \mathbf{k}_i\mathbf{i} = \mathbf{k}_i\mathbf{t}\mathbf{e}^{-ct}\qquad [\mathbf{k}_i = \mathbf{k}\pi (2\mathbf{R})^2 ]}\\ & {\therefore \mathbf{E} = -\left(\frac{\mathbf{d}\phi}{\mathbf{d}t}\right) = -\mathbf{k}_i\mathbf{e}^{-ct} + \mathbf{k}_i\mathbf{c}\mathbf{t}\mathbf{e}^{-ct}}\\ & {\qquad = -\mathbf{k}_i\mathbf{e}^{-ct}(1 - \mathbf{c}t)}\\ & {\therefore \mathbf{Induced~current~}\mathbf{l} = \frac{\mathbf{E}}{\mathbf{r}} = -\mathbf{k}_i\mathbf{e}^{-ct}(1 - \mathbf{c}t)} \end{array

1. For point charges -q, +q, +q and -q are placed on y-axis at y = -2d, y = -d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as:

E∝1D3E∝D31?E∝1DE∝D1?E∝1D4E∝D41?E∝1D2E∝D21?

Answer (3)

[image]

E‾=2qD4π?0(D2+4y2)32−2qD4π?0(D2+y2)32(−vex?)E=4π?0?(D2+4y2)23?2qD?−4π?0?(D2+y2)23?2qD?(−vex?)?E‾?=2qD4π?0D3(1(1+(2yD)2)32−1(1+(yD)2)32)?E?=4π?0?D32qD??(1+(D2y?)2)23?1?−(1+(Dy?)2)23?1??⇒E‾=2q4π?0D2(1−324y2D2−1+32y2D2)=9qy24π?0D4⇒E=4π?0?D22q?(1−23?D24y2?−1+23?D2y2?)=4π?0?D49qy2?

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1. The major product of the following reaction is :

[image]

Answer (2)

[image]

Acid catalysed intramolecular esterification

[image]

1. The peptide that gives positive ceric ammonium nitrate and carbylamine tests is

(1) Ser - Lys
(2) Lys - Asp
(3) Gln - Asp
(4) Asp - Gln

Answer (1)

Sol. Ceric ammonium nitrate test is given by alcohol. Only serine(ser) contain - OH group.

1. Which of the following compounds is a constituent of the polymer

ΩH−N−C−NH−CH2nΩH−N−C−NH−CH2??n?

(1) Formaldehyde
(2) Ammonia
(3) Methylamine
(4) N-Methyl urea

Answer (1)

Sol. ΩH−N−C−NH−CH2nΩH−N−C−NH−CH2??n? - urea-formaldehyde resin Monomers : Urea and formaldehyde

1. During compression of a spring the work done is 10kJ10kJ and 2kJ2kJ escaped to the surroundings as heat. The change in internal energy, ΔUΔU (in kJ) is

(1) 12
(2) -12
(3) 8
(4) -8

Answer (3)

Sol. w=10kJw=10kJ

q=−2kJq=−2kJΔU=q+w=10−2=8kJΔU=q+w=10−2=8kJ

1. What would be the molality of 20%20% (mass/mass) aqueous solution of KI? (molar mass of KI=166gmol−1KI=166gmol−1

(1) 1.48
(2) 1.51
(3) 1.08
(4) 1.35

Answer (2)

Sol. 20%20% W/W KI solution

i.e. 100g100g solution contains 20gKI20gKI

Mass of solvent =100−20=80g=100−20=80g

ΩMolality=20×1000166×80ΩMolality?=166×8020×1000? Ωi=1.51molarΩi?=1.51molar

1. A solution of Ni(NO3)2Ni(NO3?)2? is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

(1) 0.20

(2) 0.15

(3) 0.10

(4) 0.05

Answer (4)

Sol. 0.1 F of electricity is passed through Ni(NO3)2Ni(NO3?)2? solution

Amount of Ni deposited =0.1eq=0.1eq

Moles =0.12=0.05=20.1?=0.05

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1. HF has highest boiling point among hydrogen halides, because it has (1) Strongest hydrogen bonding (2) Lowest dissociation enthalpy (3) Strongest van der Waals' interactions (4) Lowest ionic character Answer (1) Sol. HF has highest boiling point among the hydrogen halides due to strong H- bonding between HF molecules. 8. The one that is not a carbonate ore is (1) Bauxite (2) Calamine (3) Siderite (4) Malachite Answer (1) Sol. Bauxite →AlOx(OH)3−2x→AlOx?(OH)3?−2x Calamine →ZnCO3→ZnCO3? Siderite →FeCO3→FeCO3? Siderite →FeCO3→FeCO3? Malachite →CuCO3⋅Cu(OH)2→CuCO3?⋅Cu(OH)2? 9. Noradrenaline is a / an (1) Neurotransmitter (2) Antihistamine (3) Antacid (4) Antidepressant Answer (1) Sol. Noradrenaline is neurotransmitter. 10. Among the following species, the diamagnetic molecule is (1) CO (2) NO (3) O2O2? (4) B2B2? Answer (1) Sol. Molecule No. of unpaired electrons NO 1 CO Zero O2O2? 2 B2B2? 2 Diamagnetic species is CO 11. The correct statements among I to III regarding group 13 element oxides are, (1) Boron trioxide is acidic. (II) Oxides of aluminium and gallium are amphoteric. (III) Oxides of indium and thallium are basic. (1) (II) and (III) only (2) (I) and (II) only (3) (I), (II) and (III) (4) (I) and (III) only

Answer (3)

Sol. B2O3B2?O3? is an acidic oxide

Al2O3Al2?O3? and Ga2O3Ga2?O3? are amphoteric oxide

In2O3In2?O3? and Ti2OTi2?O are basic oxide

1. The major products A and B for the following reactions are, respectively

[image]

Answer (4)

[image]

1. Increasing order of reactivity of the following compounds for SN1SN?1 substitution is

[image]

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1. B<C<A<DB<C<A<D (2) B<C<D<AB<C<D<A (3) B<A<D<CB<A<D<C (4) A<B<D<CA<B<D<C Answer (3) Sol. SN1SN?1 reaction proceeds via formation of carbocation.

[image]

On comparing (A) and (B), in (A) there is formation of tertiary carbocation CH3−CCH3?−C after rearrangement while (B) is primary. So, C)>D)>A)>B)C)>D)>A)>B)

1. Which of the following potential energy (PE) diagrams represents the SN1SN?1 reaction?

[image]

Answer (4)

Sol. In SN1SN?1 reaction, formation of carbocation (1st step) is rate determining step (RDS) Correct graph is given in option- 4.

Correct graph is given in option- 4.

1. Molal depression constant for a solvent is 4.0Kkgmol−14.0Kkgmol−1 . The depression in the freezing point of the solvent for 0.03molkg−10.03molkg−1 solution of K2SO4K2?SO4? is

Assume complete dissociation of the electrolyte)

(1) 0.36K0.36K (2) 0.18K0.18K (3) 0.12K0.12K (4) 0.24K0.24K

Answer (1)

Sol. K2SO4?2K++SO42−K2?SO4??2K++SO42−? i(Van't Hoff Factor) =3=3 ∴ΔTf=iKfm∴ΔTf?=iKf?m =3×4×0.03=3×4×0.03 =0.36K=0.36K

1. Consider the given plot of enthalpy of the following reaction between A and B.

A+B?C+DA+B?C+D

Identify the incorrect statement.

[image]

(1) Activation enthalpy to form C is 5kJmol−15kJmol−1 less than that to form D

(2) D is kinetically stable product

(3) Formation of A and B from C has highest enthalpy of activation

(4) C is the thermodynamically stable product

Answer (1)

Sol. Activation enthalpy to form C is 5kJ5kJ more than that to form D.

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1. In an acid-base titration, 0.1M HCl0.1M HCl solution was added to the NaOH solution of unknown strength. Which of the following correctly shows the change of pH of the titration mixture in this experiment?

[image]

(1) (A)
(2) (C)
(3) (B)
(4) (D)

Answer (1)

Sol. The pH of NaOH is more than 7 and during the titration it decreases so graph (1) is correct

1. Hinsberg's reagent is

 

Answer (1)

Sol. Hinsberg's reagent is benzenesulphonyl chloride

[image]

1. The layer of atmosphere between 10km10km to 50km50km above the sea level is called as

(1) Stratosphere
(2) Mesosphere
(3) Thermosphere
(4) Troposphere

Answer (1)

Sol. Between 10- 50 km above sea level lies stratosphere.

1. Assertion : For the extraction of iron, haematite ore is used.

Reason : Haematite is a carbonate ore of iron.

(1) Only the reason is correct

(2) Only the assertion is correct

(3) Both the assertion and reason are correct and the reason is the correct explanation for the assertion
(4) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion

Answer (2)

Sol. For the extraction of iron, haematite ore in used.

Haematite=Fe2O3Haematite=Fe2?O3?

1. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as

P=RTV−batT.P=V−bRT?atT.

Here, b is the van der Waal's constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs P?

(1) Kr
(2) Ar
(3) Xe
(4) Ne

Answer (3)

Sol. P=RTVm−bSol. P=Vm?−bRT? ⇒PVm−Pb=RT⇒PVm?−Pb=RT ⇒PVmRT=1+PbRT⇒RTPVm??=1+RTPb? ⇒Z=1+PbRT⇒Z=1+RTPb?

Slope of Z vs P curve (straight line) = bRTRTb?

Higher the value of b, more steep will be the curve and b≈b≈ size of gas molecules

1. The amorphous form of silica is

(1) Quartz
(2) Tridymite
(3) Kieselguhr
(4) Cristobalite

Answer (3)

Sol. Quartz, tridymite and cristobalite are crystalline forms of silica.

Kieselguhr is an amorphous form of silica.

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1. The correct statements among I to III are

(I) Valence bond theory cannot explain the color exhibited by transition metal complexes. (II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes. (III) Valence bond theory cannot distinguish ligands as weak and strong field ones. (1) (II) and (III) only (2) (I), (II) and (III) (3) (I) and (II) only (4) (I) and (III) only

Answer (4)

Sol. Valence bond theory cannot predict quantitatively the magnetic properties of transition metal complex.

1. In the following reaction

carbons compound ++ MeOH HCI Rate of the reaction is the highest for

(1) Acetone as substrate and methanol in excess
(2) Propanal as substrate and methanol in stoichiometric amount
(3) Propanal as substrate and methanol in excess
(4) Acetone as substrate and methanol in stoichiometric amount

Answer (3)

[image]

Generally, aldehydes are more reactive than ketones in nucleophilic addition reactions.

Rate of reaction with alcohol to form acetal and ketal is

[image]

1. The structures of beryllium chloride in the solid state and vapour phase, respectively, are

(1) Chain and dimeric
(2) Dimeric and dimeric
(3) Dimeric and chain
(4) Chain and chain

Answer (1)

Sol. BeCl2BeCl2? in vapour phase exist as dimer (below 1200 K temperature)

BeCl2BeCl2? in solid state has chain structure.

1. p-Hydroxybenzonphenone upon reaction with bromine in carbon tetrachloride gives

[image]

Answer (4)

[image]

Product will formed as per - OH group +M+M group)

1. The maximum possible denticities of a ligand given below towards a common transition and inner-transition metal ion, respectively, are

[image]

(1) 6 and 8
(2) 8 and 6
(3) 8 and 8
(4) 6 and 6

Answer (1)

Sol. The maximum possible denticities of the given ligand towards transition metal ion is 6 and towards inner transition metal ion is 8.

1. 10 mL of 1 mM surfactant solution forms a monolayer covering 0.24cm20.24cm2 on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

(1) 2.0 pm
(2) 2.0 nm
(3) 0.1 nm
(4) 1.0 pm

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1

Sol. No. of surfactant molecule 6×1023×101000×10−36×1023×100010?×10−3 ⇒6×1018⇒6×1018 molecule Let edge length =a=a cm Total surface area of surfactant =6×1018=6×1018 a² ⇒0.24⇒0.24 a=2×10−10a=2×10−10 cm =2=2 pm

1. Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a0a0? )

(1) The probability density of finding the electron is maximum at the nucleus
(2) The electron can be found at a distance 2a02a0? from the nucleus
(3) The magnitude of the potential energy is double that of its kinetic energy on an average
(4) The total energy of the electron is maximum when it is at a distance a0a0? from the nucleus

Answer (4)

Sol. The total energy of the electron is minimum when it is at a distance a0a0? from the nucleus for 1 s orbital.

1. The maximum number of possible oxidation states of actinoides are shown by

(1) Berkelium (Bk) and californium (Cf)
(2) Neptunium (Np) and plutonium (Pu)
(3) Actinium (Ac) and thorium (Th)
(4) Nobelium (No) and lawrencium (Lr)

Answer (2)

Sol. Actinoids Oxidation state shown

Th +4 Ac +3 Pu +3,+4,+5,+6,+7 Np +3,+4,+5,+6,+7 Bk +3,+4 Cm +3,+4 Lr +3

Maximum oxidation state is shown by (Np and Pu)

PART-C:MATHEMATICS

1. Two poles standing on a horizontal ground are of heights 5m5m and 10m10m respectively. The line joining their tops makes an angle of 15?15? with the ground. Then the distance (in m) between the poles, is:

5(2+3)5(2+3?) 5(3+1)5(3?+1)

Answer (1)

Sol.

[image]

=5(2+3)=5(2+3?)

1. If the system of equations 2x+3y−z=02x+3y−z=0 x+ky−2z=0x+ky−2z=0 and 2x−y+z=02x−y+z=0 has a non-trivial solution (x,y,z)(x,y,z) , then xy+yz+zx+kyx?+zy?+xz?+k is equal to:

(1)12(2)−4(1)21?(2)−4

Answer (1)

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2 3 -1 Sol. Δ=0⇒?23−11k−22−11?=0⇒?k=922−11?Δ=0⇒?212?3k−1?−1−21??=0⇒?k2?=−1?29?1?? Equations are 2x+3y−z=02x+3y−z=0 ...(i) 2x−y+z=02x−y+z=0 ...(ii) 2x+9y−4z=02x+9y−4z=0 ...(iii) By (i) - (ii) ?2y=zz=−4x??2y=zz=−4x?? ?2x+y=0xy+yz+k=−12+12−4+92??2x+y=0yx?+zy?+k=2−1?+21?−4+29??? =12=21? 3. If m is chosen in the quadratic equation (m2+1)x2−3x+(m2+1)2=0(m2+1)x2−3x+(m2+1)2=0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: (1) 8383? (2) 105105? (3) 4343? (4) 8585? Answer (4) Sol. Sum of roots =3m2+1=m2+13? For maximum m=0m=0 Hence equation becomes x2−3x+1=0x2−3x+1=0 α+β=3,αβ=1,?α−β?=5α+β=3,αβ=1,?α−β?=5? ?α3−β3?=?(α−β)(α2+β2+αβ)?=5(9−1)=85?α3−β3?=?(α−β)(α2+β2+αβ)?=5?(9−1)=85? 4. The vertices B and C of a ΔABCΔABC lie on the line, x+23=y−10=z43x+2?=0y−1?=4z? such that BC=5BC=5 units. Then the area (in sq. units) of this triangle, given that the point A(1,−1,2)A(1,−1,2) , is: (1) 517517? (2) 3434? (3) 6 (4) 234234? Answer (2) Sol.

[image]

Area of ΔABC=12×BC×ADΔABC=21?×BC×AD

Given BC=5BC=5 so we need perpendicular distance of A from line BC.

Let a point D on BC=(3λ−2,1,4λ)BC=(3λ−2,1,4λ)

AD‾=(3λ−3)i^+2j^+(4λ−2)k^AD=(3λ−3)i^+2j^?+(4λ−2)k^

Also AD‾AD & BC‾BC should be perpendicular

AD‾⋅BC‾=0AD⋅BC=0(3λ−3)3+2(0)+(4λ−2)4=0(3λ−3)3+2(0)+(4λ−2)4=09λ−9+16λ−8=0⇒λ=17259λ−9+16λ−8=0⇒λ=2517?

Hence, D=(125,1,6825)D=(251?,1,2568?)

AD‾=(125−1)2+(2)2+(6825−2)2AD=(251?−1)2+(2)2+(2568?−2)2? =(−2425)2+4+(1825)2=(25−24?)2+4+(2518?)2? =(2425)2+4(2525)2+(1825)2=(2524?)2+4(2525?)2+(2518?)2? =576+2500+324252=252576+2500+324?? =3400252=2523400?? =34⋅1025=2345=2534⋅10??=5234?? =12×12×12×2345=34=21?×21?×21?×5234??=34?

Area of triangle =12×BC‾×AD‾=21?×BC×AD

=12×5×2345=34=21?×5×5234??=34?

1. If cos?dydx−ysin?x=6xcosdxdy?−ysinx=6x (0<x<π2)(0<x<2π?) and y(π3)=0y(3π?)=0 , then y(π6)y(6π?) is equal to :

(1)−π22(2)−π243(1)−2π2?(2)−43?π2? (3)π223(4)−π223(3)23?π2?(4)−23?π2?

Answer (4)

===== Page 16 =====

1. cosxdy- (sinx)ydx=6xdx

⇒∫d(ycos?x)=∫6xdx⇒∫d(ycosx)=∫6xdx⇒ycos?x=3x2+C⇒ycosx=3x2+CAsy(π3)=0⇒(0)×(12)=3π29+C⇒C=−π23Asy(3π?)=0⇒(0)×(21?)=93π2?+C⇒C=3−π2?⇒ycos?x=3x2−π23⇒ycosx=3x2−3π2?Fory(π6)Fory(6π?)y32=3π236−π23y23??=363π2?−3π2?3y2=−3π212⇒y=−π22323?y?=12−3π2?⇒y=23?−π2?

1. If ∫esec?x(sec?xtan?xf(x)+sec?xtan?x+sec?2x)dx∫esecx(secxtanxf(x)+secxtanx+sec2x)dx =esec?xf(x)+C=esecxf(x)+C , then a possible choice of f(x)f(x) is:

(1) secx−tanx−12(2) secx+tanx+12(1) secx−tanx−21?(2) secx+tanx+21?(3) secx+tanx−12(4) xsecx+tanx+12(3) secx+tanx−21?(4) xsecx+tanx+21?

Answer (2)

Sol.∫esec?x(sec?xtan?xf(x)+(sec?xtan?x+sec?2x))dxSol.∫esecx(secxtanxf(x)+(secxtanx+sec2x))dx =esec?xf(x)+C=esecxf(x)+C

We know that

∫eg(x)((g′(x)f(x))+f′(x))dx=eg(x)×f(x)+C∫eg(x)((g′(x)f(x))+f′(x))dx=eg(x)×f(x)+C∴f(x)=∫((sec?xtan?x)+sec?2x)dx∴f(x)=∫((secxtanx)+sec2x)dx∴f(x)=sec?x+tan?x+C∴f(x)=secx+tanx+C

1. If some three consecutive coefficients in the binomial expansion of (x+1)n(x+1)n in powers of xx are in the ratio 2:15:70, then the average of these three coefficients is:

(1) 625
(2) 964

(3) 232
(4) 227

Answer (3)

Sol. Given nCr−1:nCr:nCr+1=2:15:70nCr−1?:nCr?:nCr+1?=2:15:70

nCr−1nCr=215&nCrnCr+1=1570nCr?nCr−1??=152?&nCr+1?nCr??=7015?⇒rn−r+1=215&r+1n−r=314⇒n−r+1r?=152?&n−rr+1?=143?⇒15r=2n−2r+2&14r+14=3n−3r⇒15r=2n−2r+2&14r+14=3n−3r⇒17r=2n+2&17r=3n−14⇒17r=2n+2&17r=3n−14i.e.,2n+2=3n−14⇒n=16&r=2i.e.,2n+2=3n−14⇒n=16&r=2Mean=16C1+16C2+16C33=16+120+5603Mean=316C1?+16C2?+16C3??=316+120+560?=6963=232=3696?=232

1. A rectangle is inscribed in a circle with a diameter lying along the line 3y=x+73y=x+7 . If the two adjacent vertices of the rectangle are (−8,5)(−8,5) and (6,5)(6,5) , then the area of the rectangle (in sq. units) is:

(1) 56
(2) 84

(3) 72
(4) 98

Answer (2)

Sol. Given situation

[image]

Perpendicular bisector of AB will pass from centre.

Equation of perpendicular bisector x=−1x=−1

Hence centre (−1,2)(−1,2)

Let

D=(α,β)⇒α+62=−1&β+52=2D=(α,β)⇒2α+6?=−1&2β+5?=2 α=−8&β=−1D=(−8,−1)α=−8&β=−1D=(−8,−1)

?AD?=6&?AB?=14?AD?=6&?AB?=14Area=6×14=84Area=6×14=84

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1. If a unit vector a^a^ makes angles π33π? with i^i^, π44π? with j^j^? and θ∈(0,π)θ∈(0,π) with k^k^, then a value of θθ is :

(1) 5π12125π?
(2) 2π332π?

(3) π44π?
(4) 5π665π?

Answer (2)

Sol. Let cos?α,cos?β,cos?γcosα,cosβ,cosγ be direction cosines of a^a^

Hence, by given data

cos?α=cos?π3,cos?β=cos?π4&cos?γ=cos?θcosα=cos3π?,cosβ=cos4π?&cosγ=cosθ∴cos?2π3+cos?2π4+cos?2θ=1∴cos23π?+cos24π?+cos2θ=1cos?2θ=14⇒cos?θ=±12,θ=π3or2π3cos2θ=41?⇒cosθ=±21?,θ=3π?or32π?

1. If the function f(x)={a?π−x?+1,x≤5b?x−π?+3,x>5f(x)={a?π−x?+1,b?x−π?+3,?x≤5x>5? is continuous at x = 5, then the value of a - b is:

(1) 2π−5π−52?
(2) −2π+5π+5−2?

(3) 2π+5π+52?
(4) 25−π5−π2?

Answer (4)

Sol. L.H.L. lim?x→5b?π−5?+3=(5−π)b+3limx→5?b?π−5?+3=(5−π)b+3

f(5) = R.H.L. lim?x→5a?5−π?+1=a(5−π)+1limx→5?a?5−π?+1=a(5−π)+1

For continuity LHL = RHL

(5−π)b+3=(5−π)a+1(5−π)b+3=(5−π)a+1⇒2=(a−b)(5−π)⇒2=(a−b)(5−π)⇒a−b=25−π⇒a−b=5−π2?

1. Let z ∈ C be such that |z| < 1. If ω=5+3z5(1−z)ω=5(1−z)5+3z?, then

(1) 5Re(ω)>45Re(ω)>4
(2) 5Re(ω)>15Re(ω)>1

(3) 4Im(ω)>54Im(ω)>5
(4) 5Im(ω)<15Im(ω)<1

Answer (2)

Sol. ω=5+3z5−5z⇒5ω−5ωz=5+3zω=5−5z5+3z?⇒5ω−5ωz=5+3z

⇒5ω−5=z(3+5ω)⇒5ω−5=z(3+5ω)⇒z=5(ω−1)3+5ω⇒z=3+5ω5(ω−1)?

Given |z| < 1

⇒5?ω−1?<?3+5ω?⇒5?ω−1?<?3+5ω?⇒25(ωω?−ω−ω?+1)<9+25ωω?+15ω+15ω?⇒25(ωω?−ω−ω?+1)<9+25ωω?+15ω+15ω?

(using |z|² = z \bar{z})

⇒16<40ω+40ω?⇒16<40ω+40ω?⇒ω+ω?>25⇒ω+ω?>52?⇒2Re(ω)>25⇒Re(ω)>15⇒2Re(ω)>52?⇒Re(ω)>51?

1. Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

(1) 635635?
(2) 1755?17?

(3) 20552055?
(4) 1155?11?

Answer (4)

Sol. Let the plane be

P ≡ (2x + 3y + z + 5) + λ(x + y + z - 6) = 0

As the above plane is perpendicular to xy plane

⇒((2+λ)i^+(3+λ)j^+(1+λ)k^)⋅k^=0⇒((2+λ)i^+(3+λ)j^?+(1+λ)k^)⋅k^=0⇒λ=−1⇒λ=−1

P ≡ x + 2y + 11 = 0

Distance from (0, 0, 256)

?0+0+115?=115?5?0+0+11??=5?11?

1. If f(x) = [x] - [x4][4x?], x ∈ R, where [x] denotes the greatest integer function, then :

(1) lim?x→4+f(x)limx→4+?f(x) exists but lim?x→4−f(x)limx→4−?f(x) does not exist

(2) f is continuous at x = 4

(3) lim?x→4−f(x)limx→4−?f(x) exists but lim?x→4+f(x)limx→4+?f(x) does not exist

(4) Both lim?x→4−f(x)limx→4−?f(x) and lim?x→4+f(x)limx→4+?f(x) exist but are not equal

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1. Sol. L.H.L lim?x→4[x]−[x4]=3−0=3limx→4?[x]−[4x?]=3−0=3 (x<4⇒[x]=3&x4<1⇒[x4]=0)(x<4⇒[x]=3&4x?<1⇒[4x?]=0) R.H.L lim?x→4[x]−[x4]=4−1=3limx→4?[x]−[4x?]=4−1=3 (x>4⇒[x]=4&x4>1⇒[x4]=1)(x>4⇒[x]=4&4x?>1⇒[4x?]=1) f(4)=[4]−[44]=4−1=3f(4)=[4]−[44?]=4−1=3 LHL =f(4)=RHL=f(4)=RHL Hence f(x)f(x) is continuous at x=4x=4 14. The common tangent to the circles x2+y2=4x2+y2=4 and x2+y2+6x+8y−24=0x2+y2+6x+8y−24=0 also passes through the point : (1) (-6,4) (2) (-4,6) (3) (4, -2) (4) (6, -2) Answer (4) Sol. In given situation dc1c2=?r1−r2?dc1?c2??=?r1?−r2?? Common tangent S1−S2=0S1?−S2?=0 6x+8y−20=0⇒3x+4y−10=06x+8y−20=0⇒3x+4y−10=0 Hence (6, -2) lies on it 15. The domain of the definition of the function f(x)=14−x2+log?10(x3−x)f(x)=4−x21?+log10?(x3−x) is : (1) (-1,0) ∪(1,2)∪(3,∞)∪(1,2)∪(3,∞) (2) (-1,0) ∪(1,2)∪(2,∞)∪(1,2)∪(2,∞) (3) (1,2) ∪(2,∞)∪(2,∞) (4) (-2, -1) ∪(−1,0)∪(2,∞)∪(−1,0)∪(2,∞) Answer (2) Sol. For domain denominator ≠0?=0 4−x2≠0⇒x≠±24−x2?=0⇒x?=±2 (1) and x3−x>0x3−x>0 ⇒x(x−1)(x+1)>0⇒x(x−1)(x+1)>0 ∴?x∈(−1,0)∪(1,∞)∴x??∈(−1,0)∪(1,∞) (2) Hence domain is intersection of (1) & (2) i.e. x∈(−1,0)∪(1,2)∪(2,∞)x∈(−1,0)∪(1,2)∪(2,∞) 16. The sum of the series 1+2×3+3×5+4×7+…1+2×3+3×5+4×7+… upto 11th11th term is : (1) 916 (2) 946 (3) 945 (4) 915 Answer (2) Sol. 1+2.3+3.5+4.7+…1+2.3+3.5+4.7+… Lets break the sequence as shown 1+(2.3+3.5+4.7+…)S1+S(2.3+3.5+4.7+…)? We find s10=∑n=110(n+1)(2n+1)s10?=∑n=110?(n+1)(2n+1) Σ=∑n=110(2n2+3n+1)Σ=∑n=110?(2n2+3n+1) Σ=2n(n+1)(2n+1)6+3n(n+1)2+n(n=10)Σ=62n(n+1)(2n+1)?+23n(n+1)?+n(n=10) Σ=2.10.11.216+3.10.112+10Σ=62.10.11.21?+23.10.11?+10 Σ=770+165+10=945Σ=770+165+10=945 Hence required sum =1+945=946=1+945=946 17. The area (in sq. units) of the smaller of the two circles that touch the parabola, y2=4xy2=4x at the point (1, 2) and the x- axis is : (1) 4π(3+2)4π(3+2?) (2) 8π(2−2)8π(2−2?) (3) 8π(3−22)8π(3−22?) (4) 4π(2−2)4π(2−2?) Answer (3)

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[image]

The circle and parabola will have common tangent at P(1, 2)

∴ Equation of tangent to parabola

≡y×(2)=4(x+1)2⇒2y=2x+2≡y×(2)=42(x+1)?⇒2y=2x+2y=x+1y=x+1

Let equation of circle be (using family of circles)

(x−x1)2+(y−y1)2+λT=0(x−x1?)2+(y−y1?)2+λT=0⇒c≡(x−1)2+(y−2)2+λ(x−y+1)=0⇒c≡(x−1)2+(y−2)2+λ(x−y+1)=0

Also circle touches x-axis ⇒ y-coordinate of centre = radius

⇒c≡x2+y2+(λ−2)x+(−λ−4)y+(λ+5)=0⇒c≡x2+y2+(λ−2)x+(−λ−4)y+(λ+5)=0λ+42=(λ−22)2+(−λ−42)2−(λ+5)2λ+4?=(2λ−2?)2+(2−λ−4?)2−(λ+5)?⇒λ2−4λ+44=λ+5⇒λ2−4λ+4=4λ+20⇒4λ2−4λ+4?=λ+5⇒λ2−4λ+4=4λ+20⇒λ2−8λ−16=0⇒λ2−8λ−16=0⇒λ=8±64+642⇒λ=28±64+64??=4±42=4±42?λ=4−42 (Other value forms bigger circle)λ=4−42? (Other value forms bigger circle)

Hence centre of circle (22−2,4−22)(22?−2,4−22?)

Radius =4−22=4−22?

Area =π(4−22)2=8π(3−22)=π(4−22?)2=8π(3−22?)

1. The total number of matrices

A=(02y12xy−12x−y1),(x,y∈R,x≠y)A=?02x2x?2yy−y?1−11??,(x,y∈R,x?=y)

for which ATA=3I3ATA=3I3? is :

(1) 6
(2) 3

(3) 4
(4) 2

Answer (3)

Sol. ATA=[02x2x2yy−y1−11][02y12xy−12x−y1]=3IATA=?02y1?2xy−1?2x−y1???02x2x?2yy−y?1−11??=3I

=[8x20006y20003]=[300030003]=?8x200?06y20?003??=?300?030?003??⇒8x2=3,6y2=3⇒8x2=3,6y2=3x=±38,y=±12x=±83??,y=±21??

Total combinations of (x, y) = 2 × 2 = 4

1. If p⇒(q∨r)p⇒(q∨r) is false, then truth values of p, q, r are respectively :

(1) T, F, F
(2) F, F, F

(3) T, T, F
(4) F, T, T

Answer (1)

Sol. For p⇒q∨rp⇒q∨r to be F

r must be F & p → q must be F

for p → q to be F p → T & q → F

p, q, r = T, F, F

1. The mean and the median of the following ten numbers in increasing order

10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then yxxy? is equal to :

(1) 7337?

(2) 8338?

(3) 7227?

(4) 9449?

Answer (1)

Sol. Mean =∑xin=x+y+30010=42⇒x+y=120=n∑xi??=10x+y+300?=42⇒x+y=120

Median =T5+T62=35=34+x2⇒x=36&y=84=2T5?+T6??=35=234+x?⇒x=36&y=84

Hence yx=8436=73xy?=3684?=37?

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1. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is :

(1) 157
(2) 225
(3) 262
(4) 190

Answer (4)

Sol. Balls used in equilateral triangle n(n+1)22n(n+1)?

Here side of equilateral triangle has n- balls

No. of balls in each side of square is =(n−2)=(n−2)

Givenn(n+1)2+99=(n−2)2Given2n(n+1)?+99=(n−2)2⇒n2+n+198=2n2−8n+8⇒n2+n+198=2n2−8n+8⇒n2−9n−190=0⇒n2−9n−190=0⇒n2−19n+10n−190=0⇒n2−19n+10n−190=0⇒(n−19)(n+10)=0⇒(n−19)(n+10)=0⇒n=19⇒n=19

Balls used to form triangle

n(n+1)2=19×202=1902n(n+1)?=219×20?=190

1. Two newspapers A and B are published in a city. It is known that 25%25% of the city population reads A and 20%20% reads B while 8%8% reads both A and B. Further, 30%30% of those who read A but not B look into advertisements and 40%40% of those who read B but not A also look into advertisements, while 50%50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisements is :

(1) 13.9

(2) 13

(3) 12.8

(4) 13.5

Answer (1)

[image]

nA only) =25−8=17%=25−8=17%

nB only) =20−8=12%=20−8=12%

% of people from A only who read advertisement =17×0.3=5.1%=17×0.3=5.1%

% of people from B only who read advertisement =12×0.4=4.8%=12×0.4=4.8%

% of people from A&B both who read advertisement =8×0.5=4%=8×0.5=4%

Total % of people who read advertisement

=5.1+4.8+4=13.9%=5.1+4.8+4=13.9%

1. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is

tan?−1(12)tan−1(21?) Water is poured into it at a constant rate of 5 cubic metre per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m10m is :

2π(2)115ππ2?(2)15π1? 110π(4)15π10π1?(4)5π1?

Answer (4)

[image]

Sol.

dvdt=5m3/mindtdv?=5m3/min

V=13πr2h…(i)(where r is radius and h is V=31?πr2h…(i)(where r is radius and h is  height at any time)height at any time)

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[image]

Also tan?θ=rh=12⇒h=2r⇒dhdt=2drdt…(ii)tanθ=hr?=21?⇒h=2r⇒dtdh?=dt2dr?…(ii)

Differentiate eq?. (i), we get

dVdt=13(2πrdrdth+πr2dhdt)=13(100π12+25π)dhdtdtdV?=31?(2πrdtdr?h+πr2dtdh?)=31?(100π21?+25π)dtdh?

at h = 10, r = 5

5=75π3dhdt5=375π?dtdh?⇒dhdt=15πm/min⇒dtdh?=5π1?m/min

1. The value of the integral

∫01xcot?−1(1−x2+x4)dx is∫01?xcot−1(1−x2+x4)dx is

(1) π4−log?e24π?−loge?2
(2) π2−log?e22π?−loge?2

(3) π2−12log?e22π?−21?loge?2
(4) π4−12log?e24π?−21?loge?2

Answer (4)

Sol. ∫01xcot?−1(1−x2+x4)dx=∫01xtan?−1(11+x4−x2)∫01?xcot−1(1−x2+x4)dx=∫01?xtan−1(1+x4−x21?)

⇒∫01xtan?−1(x2−(x2−1)1+x2(x2−1))dx⇒∫01?xtan−1(1+x2(x2−1)x2−(x2−1)?)dx⇒∫01xtan?−1x2dx−∫01xtan?−1(x2−1)dx⇒∫01?xtan−1x2dx−∫01?xtan−1(x2−1)dx

Put x2=t⇒2xdx=dtx2=t⇒2xdx=dt, (For 1st integration)
put x2−1=k⇒2xdx=dkx2−1=k⇒2xdx=dk (For 2nd integration)

⇒12∫01tan?−1tdt−12∫−10tan?−1kdk⇒21?∫01?tan−1tdt−21?∫−10?tan−1kdk⇒12(∫01tan?−1tdt−∫−10tan?−1kdk)⇒21?(∫01?tan−1tdt−∫−10?tan−1kdk)⇒12([ttan?−1t]01−∫01t1+t2dt−[ktan?−1k]−10+∫−10k1+k2dk)⇒21?([ttan−1t]01?−∫01?1+t2t?dt−[ktan−1k]−10?+∫−10?1+k2k?dk)⇒12(π4−(12ln?(1+t2))01−12(0−π4−(12ln?(1+k2))−10))⇒21?(4π?−(21?ln(1+t2))01?−21?(0−4π?−(21?ln(1+k2))−10?))⇒(π8−14ln?2)−(−π8−14ln?2)⇒(8π?−41?ln2)−(−8π?−41?ln2)⇒π4−12ln?2⇒4π?−21?ln2

1. The area (in sq. units) of the region

A={(x,y):y22≤x≤y+4} isA={(x,y):2y2?≤x≤y+4} is

(1) 18

(2) 16

(3) 533353?

(4) 30

Answer (1)

[image]

Sol.

Hence area =∫−24xdy=∫−24?xdy

=∫−24(y+4−y22)dy=∫−24?(y+4−2y2?)dy=y22+4y−y36?−24=(8+16−646)−(2−8+86)=2y2?+4y−6y3??−24?=(8+16−664?)−(2−8+68?)=(24−323)−(−6+43)=403+143=543=18=(24−332?)−(−6+34?)=340?+314?=354?=18

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1. If the two lines x+(a−1)y=1x+(a−1)y=1 and 2x+a2y=12x+a2y=1 (a ∈ R - {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is

Answer (1)

Sol. For perpendicular m1m2=−1m1?m2?=−1

Hence lines are x−2y=1x−2y=1 and 2x+y=12x+y=1

Intersection point (35,−15)(53?,−51?)

Distance from origin 925+125=1025=25259?+251??=2510??=52??

1. The value of sin?10?sin?30?sin?50?sin?70?sin10?sin30?sin50?sin70? is

 (1) 118 (3) 116 (4) 136(2) (1) 181? (3) 161? (4) 361??(2)

Answer (3)

Sol. sin?(60?+A)sin(60?+A) sin?(60?−A)sin?A=14sin?3Asin(60?−A)sinA=41?sin3A

Hence, sin?10?sin?50?sin?70?sin10?sin50?sin70?

=sin?10?sin?(60?−10?)sin?(60?+10?)=14sin?30?=sin10?sin(60?−10?)sin(60?+10?)=41?sin30?

Hence, sin?10?sin?30?sin?50?sin?70?=14sin?2?30?sin10?sin30?sin50?sin70?=41?sin2?30?

1. If the tangent to the parabola y2=xy2=x at a point (α,(α, β)β) (β>0)(β>0) is also a tangent to the ellipse, x2+2y2=1x2+2y2=1 then αα is equal to

(1)2−1(1)2?−1 (2)2+1(2)2?+1 (3)22+1(3)22?+1 (4)22−1(4)22?−1

Answer (2)

Sol. Let tangent in terms of m to parabola and ellipse

i.e y=mx+14m for parabola at point (14m2,−12m)i.e y=mx+4m1? for parabola at point (4m21?,2m−1?)

and y=mx±m2+12y=mx±m2+21?? for ellipse on comparing

⇒14m=±m2+12⇒116m2=m2+12⇒4m1?=±m2+21??⇒16m21?=m2+21?⇒16m4+8m2−1=0⇒16m4+8m2−1=0m2=−8±64+642(16)=−8±822(16)=2−14m2=2(16)−8±64+64??=2(16)−8±82??=42?−1?α=14m2=12−1=2+1α=4m21?=2?−11?=2?+1

1. If f:R→Rf:R→R is a differentiable function and

f(2)=6,thenlim?x→2∫6f(x)2tdtf(2)=6,thenx→2lim?∫6f(x)?2tdt

(1)0(2)2f(2)(1)0(2)2f(2) (3)12f(2)(4)24f(2)(3)12f(2)(4)24f(2)

Answer (3)

Sol. Using L' Hospital rule and Leibnitz theorem

lim?x→22f(x)f′(x)−01limx→2?12f(x)f′(x)−0? 2f(2)f′(2)=12f(2)2f(2)f′(2)=12f(2)

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1. If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th11th term is

(1)- 36 (2)25 (3)- 25 (4)- 35

Answer (3)

Sol. Let terms be a - d, a, a + d

⇒3a=33⇒11⇒3a=33⇒11

Product of terms

(a−d)a(a+d)=11(121−d2)=1155(a−d)a(a+d)=11(121−d2)=1155⇒121−d2=105⇒d=±4⇒121−d2=105⇒d=±4

if d=4d=4

if d=−4d=−4

T1=15T2=11T3=7⇒T11=T1+10d=15+10(−4)=−25T1?=15T2?=11T3?=7?⇒T11?=T1?+10d=15+10(−4)=−25

JEE MAINS 2019 PHASE 2 , 9 APRIL 2019 EVENING SHIFT

JEE MAINS 2019 PHASE 2 SOLVED QUESTION PAPER

JEE MAINS PREVIOUS YEAR QUESTION PAPER

NOTE : For image and diagram and exact form of question paper kindly refer Page No 49 to 71 from given below pdf (download link given below )

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