FINAL JEE MAIN EXAM AUGUST 2021 , 31/08/2021 / EVENING SESSION 

PHYSICS
SECTION-A

1. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50×103kg50×103kg. The inner and outer radii of each column are 50 cm50 cm and 100 cm100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y=2.0×1011PaY=2.0×1011Pa, g=9.8m/s2g=9.8m/s2]
   (1) 3.60×10−83.60×10−8
   (2) 2.60×10−72.60×10−7
   (3) 1.87×10−31.87×10−3
   (4) 7.07×10−47.07×10−4
   Official Ans. by NTA (2)
   Sol. Force on each column =mg4=4mg?
   Strain =mg4AY=4AYmg?
   =50×103×9.84×π(1−0.25)×2×1011=4×π(1−0.25)×2×101150×103×9.8?
   =2.6×10−7=2.6×10−7

2. A current of 1.5A1.5A is flowing through a triangle, of side 9cm9cm each. The magnetic field at the centroid of the triangle is :
   (Assume that the current is flowing in the clockwise direction.)
   (1) 3×10−7T3×10−7T outside the plane of triangle
   (2) 23×10−7T23?×10−7T outside the plane of triangle
   (3) 23×10−5T23?×10−5T inside the plane of triangle
   (4) 3×10−5T3×10−5T inside the plane of triangle
   Official Ans. by NTA (4)
   Sol.
   B=3[μ014π(sin?60?+sin?60?)]B=3[4πμ01??(sin60?+sin60?)]
   tan?60?=?/2rtan60?=r?/2?
   Where r=9×10−223Mr=23?9×10−2?M
   ∴B=3×10−5T∴B=3×10−5T
   Current is flowing in clockwise direction so, B?B? is inside plane of triangle by right hand rule.

3. A system consists of two identical spheres each of mass 1.5kg1.5kg and radius 50 cm50 cm at the end of light rod. The distance between the centres of the two spheres is 5m5m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?
   (1) 18.75kgm218.75kgm2
   (2) 1.905×105kgm21.905×105kgm2
   (3) 19.05kgm219.05kgm2
   (4) 1.875×105kgm21.875×105kgm2
   Official Ans. by NTA (3)
   M=1.5kg,r=0.5m,d=52mM=1.5kg,r=0.5m,d=25?m
   I=2(25Mr2+Md2)I=2(52?Mr2+Md2)
   =19.05kgm2=19.05kgm2

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1. Statement I: Two forces (P?+Q?)(P?+Q??) and (P?−Q?)(P?−Q??) where P?⊥Q?P?⊥Q?? when act at an angle θ1θ1? to each other, the magnitude of their resultant is 3(P2+Q2)3(P2+Q2)?, when they act at an angle θ2θ2?, the magnitude of their resultant becomes 2(P2+Q2)2(P2+Q2)?. This is possible only when θ1<θ2θ1?<θ2?.
   Statement II: In the situation given above. θ1=60?θ1?=60? and θ2=90?θ2?=90?
   In the light of the above statements, choose the most appropriate answer from the options given below :-
   (1) Statement-I is false but Statement-II is true
   (2) Both Statement-I and Statement-II are true
   (3) Statement-I is true but Statement-II is false
   (4) Both Statement-I and Statement-II are false.
   Official Ans. by NTA (2)
   Sol. A?=P?+Q?A?=P?+Q??
   B?=P?−Q?B?=P?−Q??
   P?⊥Q?P?⊥Q??
   ?A??=?B??=P2+Q2?A??=?B??=P2+Q2?
   ?A?+B??=2(P2+Q2)(1+cos?θ)?A?+B??=2(P2+Q2)(1+cosθ)?
   For ?A?+B??=3(P2+Q2)?A?+B??=3(P2+Q2)?
   θ1=60?θ1?=60?
   For ?A?+B??=2(P2+Q2)?A?+B??=2(P2+Q2)?
   θ2=90?θ2?=90?

2. A free electron of 2.6eV2.6eV energy collides with a H+H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h=6.6×10−34(h=6.6×10−34 Js)
   (1) 1.45×1016MHz1.45×1016MHz
   (2) 0.19×1015MHz0.19×1015MHz
   (3) 1.45×109MHz1.45×109MHz
   (4) 9.0×1027MHz9.0×1027MHz
   Official Ans. by NTA (3)
   Sol. For every large distance P.E.=0P.E.=0 & total energy =2.6+0=2.6eV=2.6+0=2.6eV
   Finally in first excited state of HH atom total energy =−3.4eV=−3.4eV
   Loss in total energy =2.6−(−3.4)=2.6−(−3.4)
   =6eV=6eV
   It is emitted as photon
   λ=12406=206nmλ=61240?=206nm
   f=3×108206×10−9=1.45×1015Hzf=206×10−93×108?=1.45×1015Hz
   =1.45×109Hz=1.45×109Hz

3. Two thin metallic spherical shells of radii r1r1? and r2r2? (r1<r2)(r1?<r2?) are placed with their centres coinciding. A material of thermal conductivity KK is filled in the space between the shells. The inner shell is maintained at temperature θ1θ1? and the outer shell at temperature θ2(θ1<θ2)θ2?(θ1?<θ2?). The rate at which heat flows radially through the material is :-
   (1) 4πKr1r2(θ2−θ1)r2−r1r2?−r1?4πKr1?r2?(θ2?−θ1?)?
   (2) πr1r2(θ2−θ1)r2−r1r2?−r1?πr1?r2?(θ2?−θ1?)?
   (3) K(θ2−θ1)r2−r1r2?−r1?K(θ2?−θ1?)?
   (4) K(θ2−θ1)(r2−r1)4πr1r24πr1?r2?K(θ2?−θ1?)(r2?−r1?)?
   Official Ans. by NTA (1)
   Thermal resistance of spherical sheet of thickness dr and radius r is
   dR=drK(4πr2)dR=K(4πr2)dr?
   R=∫r1r2drK(4πr2)R=∫r1?r2??K(4πr2)dr?
   R=14πK(1r1−1r2)=14πK(r2−r1r1r2)R=4πK1?(r1?1?−r2?1?)=4πK1?(r1?r2?r2?−r1??)
   thermal current (i)=θ2−θ1R(i)=Rθ2?−θ1??
   i=4πKr1r2r2−r1(θ2−θ1)i=r2?−r1?4πKr1?r2??(θ2?−θ1?)

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Final JEE-Main Exam August, 2021/31-08-2021/Evening Session

1. If VAVA? and VBVB? are the input voltages (either 5V or 0V) and VoVo? is the output voltage then the two gates represented in the following circuit (A) and (B) are:-
   (A)
   (1) AND and OR Gate
   (2) OR and NOT Gate
   (3) NAND and NOR Gate
   (4) AND and NOT Gate
   Official Ans. by NTA (2)
   Sol. VA=5V⇒A=1VA?=5V⇒A=1
   VA=0V⇒A=0VA?=0V⇒A=0
   VB=5V⇒B=1VB?=5V⇒B=1
   VB=0V⇒B=0VB?=0V⇒B=0
   If A=B=0A=B=0, there is no potential anywhere here
   V0=0V0?=0
   If A=1,B=0A=1,B=0, Diode D1D1? is forward biased, here
   V0=5VV0?=5V
   If A=0,B=1A=0,B=1, Diode D2D2? is forward biased hence
   V0=5VV0?=5V
   If A=1,B=1A=1,B=1, Both diodes are forward biased hence V0=5VV0?=5V
   Truth table for Ist
   A B Output
   0 0 0
   0 1 1
   1 0 1
   1 1 1
   ∴ Given circuit is OR gate
   For IInd circuit
   VB=5V,A=1VB?=5V,A=1
   VB=0V,A=0VB?=0V,A=0
   When A=0A=0, E–B junction is unbiased there is no current through it
   ∴ V0=1V0?=1
   When A=1A=1, E–B junction is forward biased
   V0=0V0?=0
   ∴ Hence this circuit is not gate.

2. Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-
   (1) (mpme)3/2(me?mp??)3/2
   (2) mempmp?me???
   (3) mpmeme?mp???
   (4) mpmeme?mp??
   Official Ans. by NTA (3)
   Sol. Δx.Δp≥h4πΔx.Δp≥4πh?
   Δx=h4πmΔvΔx=4πmΔvh?
   v=3KTmv=m3KT??
   ΔxeΔxp=mpmeΔxp?Δxe??=me?mp???

3. A bob of mass ‘m’ suspended by a thread of length ll undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1441? times that of the bob and the length of the thread is increased by 1/3rd1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-
   (1) T
   (2) 32T23?T
   (3) 34T43?T
   (4) 43T34?T
   Official Ans. by NTA (4)
   Sol. T=2π?/gT=2π?/g?
   When bob is immersed in liquid
   mgeff=mg−Buoyantforcemgeff?=mg−Buoyantforce
   mgeff=mg−vρgmgeff?=mg−vρg (σσ = density of liquid)
   =mg−vρ4g=mg−v4ρ?g
   =mg−mg4=3mg4=mg−4mg?=43mg?
   ∴ geff=3g4geff?=43g?
   T1=2π?1geffT1?=2πgeff??1???
   ?1=?+?3=4?3,?eff=3g4?1?=?+3??=34??,?eff?=43g?
   By solving
   T1=432π?/gT1?=34?2π?/g?
   T1=4T3T1?=34T?

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1. Statement :1 If three forces F?1,F?2F?1?,F?2? and F?3F?3? are represented by three sides of a triangle and F?1+F?2=−F?3F?1?+F?2?=−F?3?, then these three forces are concurrent forces and satisfy the condition for equilibrium.
   Statement :II A triangle made up of three forces F?1,F?2F?1?,F?2? and F?3F?3? as its sides taken in the same order, satisfy the condition for translatory equilibrium.
   In the light of the above statements, choose the most appropriate answer from the options given below:
   (1) Statement-I is false but Statement-II is true
   (2) Statement-I is true but Statement-II is false
   (3) Both Statement-I and Statement-II are false
   (4) Both Statement-I and Statement-II are true.
   Official Ans. by NTA (4)
   Sol. Here F?1+F?2+F?3=0F?1?+F?2?+F?3?=0
   F?1+F?2=−F?3F?1?+F?2?=−F?3?
   Since F?net=0F?net?=0 (equilibrium)
   Both statements correct

2. If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be :
   (1) [FT−1V−1][FT−1V−1]
   (2) [FTV−1][FTV−1]
   (3) [FT2V][FT2V]
   (4) [FVT−1][FVT−1]
   Official Ans. by NTA (2)
   Sol. [M]=K[F]a[T]b[V]c[M]=K[F]a[T]b[V]c
   [Mi]=[MiLiT−2]a[Ti]b[LiT−1]c[Mi]=[MiLiT−2]a[Ti]b[LiT−1]c
   a=1,b=1,c=−1a=1,b=1,c=−1
   ∴ [M]=[FTV−1][M]=[FTV−1]

3. The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2cos?(kz−ωt)B=B0?2?i^+j^??cos(kz−ωt) ; where i^,j^i^,j^? represents unit vector along x and y-axis respectively. At t=0t=0 s, two electric charges q1q1? of 4π4π coulomb and q2q2? of 2π2π coulomb located at (0,0,πk)(0,0,kπ?) and (0,0,3πk)(0,0,k3π?), respectively, have the same velocity of 0.5ci^0.5ci^, (where c is the velocity of light). The ratio of the force acting on charge q1q1? to q2q2? is :-
   (1) 22:122?:1
   (2) 1:21:2?
   (3) 2:12:1
   (4) 2:12?:1
   Official Ans. by NTA (3)
   Sol. F?=q(V?×B?)F?=q(V?×B?)
   F?1=4π[0.5ci×B0(i^+j^2)cos?(Kπ−0)]F?1?=4π[0.5ci×B0?(2i^+j^??)cos(K−π?0)]
   F?2=2π[0.5ci×B0(i^+j^2)cos?(K3π−0)]F?2?=2π[0.5ci×B0?(2i^+j^??)cos(K−3π?0)]
   cos?π=−1,cos?3π=−1cosπ=−1,cos3π=−1
   ∴ F1F2=2F2?F1??=2

4. The equivalent resistance of the given circuit between the terminals A and B is :
   (1) 0Ω0Ω
   (2) 3Ω3Ω
   (3) 92Ω29?Ω
   (4) 1Ω1Ω
   Official Ans. by NTA (4)

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1. A mixture of hydrogen and oxygen has volume 500 cm3500 cm3, temperature 300 K300 K, pressure 400 kPa400 kPa and mass 0.76g0.76g. The ratio of masses of oxygen to hydrogen will be :-
   (1) 3:83:8
   (2) 3:163:16
   (3) 16:316:3
   (4) 8:38:3
   Official Ans. by NTA (3)
   Sol. PV=nRTPV=nRT
   400×103×500×10−6=n(253)(300)400×103×500×10−6=n(325?)(300)
   n=225n=252?
   n=n1+n2n=n1?+n2?
   225=M12+M232252?=2M1??+32M2??
   Also M1+M2=0.76gmM1?+M2?=0.76gm
   M2M1=163M1?M2??=316?

2. A block moving horizontally on a smooth surface with a speed of 40 m/s40 m/s splits into two parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s60 m/s in the same direction, then the fractional change in kinetic energy is :-
   (1) 1331?
   (2) 2332?
   (3) 1881?
   (4) 1441?
   Official Ans. by NTA (3)
   Sol. PV=nRTPV=nRT
   400×103×500×10−6=n(253)(300)400×103×500×10−6=n(325?)(300)
   n=225n=252?
   n=n1+n2n=n1?+n2?
   225=M12+M232252?=2M1??+32M2??
   Also M1+M2=0.76gmM1?+M2?=0.76gm
   M2M1=163M1?M2??=316?

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Final JEE-Main Exam August, 2021/31-08-2021/Evening Session

1. A coil is placed in a magnetic field B?B as shown below :
   A current is induced in the coil because B?B is :
   (1) Outward and decreasing with time
   (2) Parallel to the plane of coil and decreasing with time
   (3) Outward and increasing with time
   (4) Parallel to the plane of coil and increasing with time
   Official Ans. by NTA (1)
   Sol. B?B must not be parallel to the plane of coil for non zero flux and according to lenz law if B is outward it should be decreasing for anticlockwise induced current.

2. For a body executing S.H.M. :
   (a) Potential energy is always equal to its K.E.
   (b) Average potential and kinetic energy over any given time interval are always equal.
   (c) Sum of the kinetic and potential energy at any point of time is constant.
   (d) Average K.E. in one time period is equal to average potential energy in one time period.
   Choose the most appropriate option from the options given below :
   (1) (c) and (d)
   (2) only (c)
   (3) (b) and (c)
   (4) only (b)
   Official Ans. by NTA (1)
   Sol. In S.H.M. total mechanical energy remains constant and also <K.E.> = <P.E.> = 14KA241?KA2
   (for 1 time period)

3. Statement-I :
   To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RLRL?.
   Statement-II :
   To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RLRL?.
   In the light of the above statements, choose the most appropriate answer from the options given below :
   (1) Statement I is true but Statement II is false
   (2) Statement I is false but Statement II is true
   (3) Both Statement I and Statement II are false
   (4) Both Statement I and Statement II are true
   Official Ans. by NTA (4)
   Sol. To convert pulsating dc into steady dc both of mentioned method are correct.

4. If RERE? be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ‘r’ below and a height ‘r’ above the earth surface is : (Given : r<REr<RE?)
   (1) 1−rRE−r2RE2−r3RE31−RE?r?−RE2?r2?−RE3?r3?
   (2) 1+rRE+r2RE2+r3RE31+RE?r?+RE2?r2?+RE3?r3?
   (3) 1+rRE−r2RE2+r3RE31+RE?r?−RE2?r2?+RE3?r3?
   (4) 1+rRE−r2RE2−r3RE31+RE?r?−RE2?r2?−RE3?r3?
   Official Ans. by NTA (4)
   Sol. gup=g(1+rR)2gup?=(1+Rr?)2g?
   gdown=g(1−rR)gdown?=g(1−Rr?)
   gdowngup=(1−rR)(1+rR)2gup?gdown??=(1−Rr?)(1+Rr?)2
   =(1−rR)(1+2rR+r2R2)=(1−Rr?)(1+R2r?+R2r2?)
   =1+rR−r2R2−r3R3=1+Rr?−R2r2?−R3r3?

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1. A bandwidth of 6MHz6MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6kHz6kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be
   Official Ans. by NTA (500)
   Sol. Signal bandwidth =2=2 fm
   ∴ N=6MHz12kHz=6×10612×103=500N=12kHz6MHz?=12×1036×106?=500

2. A parallel plate capacitor of capacitance 200μF200μF is connected to a battery of 200V200V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be
   Official Ans. by NTA (4)
   Sol. ΔU=12(ΔC)V2ΔU=21?(ΔC)V2
   ΔU=12(KC−C)V2ΔU=21?(KC−C)V2
   ΔU=122(2−1)CV2ΔU=212?(2−1)CV2
   ΔU=12×200×10−6×200×200ΔU=21?×200×10−6×200×200
   ΔU=4JΔU=4J

3. A long solenoid with 1000 turns/m has a core material with relative permeability 500 and volume 103cm3103cm3. If the core material is replaced by another material having relative permeability of 750 with same volume maintaining same current of 0.75 A in the solenoid, the fractional change in the magnetic moment of the core would be approximately (x499)(499x?). Find the value of x.
   Official Ans. by NTA (250)
   Sol. ΔMM=Δμμ=250500=12MΔM?=μΔμ?=500250?=21?
   12=x499⇒x=25021?=499x?⇒x=250

4. A particle is moving with constant acceleration ‘a’. Following graph shows v2v2 versus x (displacement) plot. The acceleration of the particle is m/s2m/s2.
   Official Ans. by NTA (1)
   Sol. y=mx+Cy=mx+C
   v2=2010x+20v2=1020?x+20
   v2=2x+20v2=2x+20
   2vdvdx=22vdxdv?=2
   ∴ a=vdvdx=1a=vdxdv?=1

5. In a Young’s double slit experiment, the slits are separated by 0.3mm0.3mm and the screen is 1.5m1.5m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4cm2.4cm. The frequency of light used is Ωx×1014HzΩx?×1014Hz.
   Official Ans. by NTA (5)
   Sol. 8β=2.4cm8β=2.4cm
   8λΔd=2.4cmd8λΔ?=2.4cm
   8×1.5×c0.3×10−3×f=2.4×10−20.3×10−3×f8×1.5×c?=2.4×10−2
   f=5×1014Hzf=5×1014Hz

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1. The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1mm1mm. The main scale reading is 10mm10mm and 8th8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04cm0.04cm, then the radius of the bob is ×10−2cm×10−2cm.
   Official Ans. by NTA (52)
   Sol. 9 MSD =10=10 VSD
   9×1mm=109×1mm=10 VSD
   ∴ 1VSD=0.9mm1VSD=0.9mm
   LC=1MSD−1VSD=0.1mmLC=1MSD−1VSD=0.1mm
   Reading == MSR ++ VSR ×× LC
   10+8×0.1=10.8mm10+8×0.1=10.8mm
   Actual reading =10.8−0.4=10.4mm=10.8−0.4=10.4mm
   radius=d2=10.42=5.2mmradius=2d?=210.4?=5.2mm
   =52×10−2cm=52×10−2cm

2. A sample of gas with γ=1.5γ=1.5 is taken through an adiabatic process in which the volume is compressed from 1200cm31200cm3 to 300cm3300cm3. If the initial pressure is 200kPa200kPa. The absolute value of the workdone by the gas in the process == J.
   Official Ans. by NTA (480)
   Sol. ν=1.5ν=1.5
   p1ν1ν=p2ν2νp1?ν1ν?=p2?ν2ν?
   (200)(1200)1.5=P2(300)1.5(200)(1200)1.5=P2(300)1.5
   P2=200[4]3/2=1600kPaP2?=200[4]3/2=1600kPa
   ?W.D.?=p2ν2−p1ν1ν−1=(480−2400.5)=480 J ?W.D.?=ν−1p2?ν2?−p1?ν1??=(0.5480−240?)=480 J 

3. At very high frequencies, the effective impedance of the given circuit will be ΩΩ.
   Official Ans. by NTA (2)
   Sol. XL=2πfLXL?=2πfL
   f is very large
   ∴ XLXL? is very large hence open circuit.
   XC=12πfCXC?=2πfC1?
   f is very large.
   ∴ XCXC? is very small, hence short circuit.
   Final circuit
   Zeq=1+2×22+2=2Zeq?=1+2+22×2?=2

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Final JEE-Main Exam August, 2021/31-08-2021/Evening Session

1. Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ______ ×10−10×10−10 s. (Given, speed of light in vacuum =3×108m/s=3×108m/s and cos?30?=32cos30?=23??)
   Official Ans. by NTA (5)
   Sol. i=A=60?i=A=60?
   δmin?=2i−Aδmin?=2i−A
   =2×60?−60?=60?=2×60?−60?=60?
   μ=sin?−1(δmin?+A2)sin?−1(A2)μ=sin−1(2A?)sin−1(2δmin?+A?)?
   =3=3?
   Vprism=3×1083Vprism?=3?3×108?
   AP=10×10−2×32AP=10×10−2×23??
   time=5×10−23×108×3×3time=3×1085×10−2?×3?×3?
   =5×10−10sec=5×10−10sec
   Ans = 5

2. A resistor dissipates 192 J of energy in 1 s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s in ______ J.
   Official Ans. by NTA (3840)
   Sol. E=i2RtE=i2Rt
   192=16(R)(1)192=16(R)(1)
   R=12ΩR=12Ω
   E1=(8)2(12)(5)E1=(8)2(12)(5)
   =3840J=3840J

CHEMISTRY 

1. Arrange the following conformational isomers of n-butane in order of their increasing potential energy :

I II III IV

(1) II < III < IV < I
(2) I < IV < III < II
(3) II < IV < III < I
(4) I < III < IV < II

Official Ans. by NTA (4)

Sol. More stable less potential energy.
Stability order : I > III > IV > II
So
Potential energy : II > IV > III > I

1. The Eu2+ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]
   (1) 4f7 6s2
   (2) 4f6
   (3) 4f7
   (4) 4f6 6s2

Official Ans. by NTA (3)

Sol. Eu → [Xe]4f6 6s2
Eu2+ → [Xe]4f7

1. The structures of A and B formed in the following reaction are : [Ph = –C6H5]

(1) A = Ph–CO–CH2–CH2–COOH, B = Ph–CH2–CH2–CH2–COOH
(2) A = Ph–CO–CH2–CH3, B = Ph–CH(OH)–CH2–CH3
(3) A = Ph–CO–CH2–CH2–COOH, B = Ph–CH2–CH2–CH2–OH
(4) A = Ph–CO–CH2–CH3, B = Ph–CH2–CH2–CH3

Official Ans. by NTA (1)

Sol.

1. In which one of the following sets all species show disproportionation reaction ?
   (1) ClO2–, F2, MnO4– and Cr2O72–
   (2) Cr2O72–, MnO4–, ClO2– and Cl2
   (3) MnO4–, ClO2–, Cl2 and Mn3+
   (4) ClO4–, MnO4–, ClO2– and F2

Official Ans. by NTA (3)
Allen Ans. (Bonus)

Sol. No option contains all species that show disproportionation reaction.
MnO4–

1. Match List-I with List-II

List-I (Parameter) (Unit)
(a) Cell constant (i) S cm2 mol–1
(b) Molar conductivity (ii) Dimensionless
(c) Conductivity (iii) m–1
(d) Degree of dissociation (iv) Ω–1 m–1 of electrolyte

Choose the most appropriate answer from the options given below :
(1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(2) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
(3) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
(4) (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

Official Ans. by NTA (1)

Sol. Cell constant = (?/A) ⇒ Units = m–1
Molar conductivity (Λm) ⇒ Units = S m2 mole–1
Conductivity (K) ⇒ Units = S m–1
Degree of dissociation (α) → Dimensionless
(a)-(iii)
(b)-(i)
(c)-(iv)
(d)-(ii)

1. The major products A and B formed in the following reaction sequence are :

(1) A = acetanilide, B = p-bromoacetanilide
(2) A = acetanilide, B = o-bromoacetanilide
(3) A = p-aminoacetophenone, B = p-amino-m-bromoacetophenone
(4) A = p-aminoacetophenone, B = dibromo-p-aminoacetophenone

Official Ans. by NTA (2)
Sol.

1. Which of the following is NOT an example of fibrous protein ?
   (1) Keratin
   (2) Albumin
   (3) Collagen
   (4) Myosin

Official Ans. by NTA (2)

Sol. Keratin, collagen and myosin are example of fibrous protein.

1. The deposition of X and Y on ground surfaces is referred as wet and dry depositions, respectively. X and Y are :
   (1) X = Ammonium salts, Y = CO2
   (2) X = SO2, Y = Ammonium salts
   (3) X = Ammonium salts, Y = SO2
   (4) X = CO2, Y = SO2

Official Ans. by NTA (3)

Sol. Oxides of nitrogen and sulphur are acidic and settle down on ground as dry deposition. Ammonium salts in rain drops result in wet deposition.

1. For the reaction given below :

The compound which is not formed as a product in the reaction is a :
(1) compound with both alcohol and acid functional groups
(2) monocarboxylic acid
(3) dicarboxylic acid
(4) diol

Official Ans. by NTA (3)

Sol.

1. Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is :
   (1) 5.92
   (2) 0
   (3) 1
   (4) 1.73

Official Ans. by NTA (4)

1. Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Lithium salts are hydrated.
Reason (R) : Lithium has higher polarising power than other alkali metal group members.

In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both (A) and (R) are correct but (R) is NOT the correct explanation of (A).
(2) (A) is correct but (R) is not correct.
(3) (A) is not correct but (R) is correct.
(4) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Official Ans. by NTA (1)

Sol. Lithium salts are hydrated due to high hydration energy of Li+ Li due to smallest size in IA group has highest polarizing power.

1. The incorrect expression among the following is:

(1) ΔGSystem/ΔSTotal = –T (at constant P)
(2) ln K = (ΔH° – TΔS°)/RT
(3) K = e^(–ΔG°/RT)
(4) For isothermal process, w_reversible = –nRT ln(Vi/Vi)

Official Ans. by NTA (2)

Sol. Option (2) is incorrect
ΔG° = –RT ln K
ΔH° – TΔS° = –RT ln K
ln K = –[(ΔH° – ΔS°)/RT]

1. Which one of the following statements is incorrect ?
   (1) Atomic hydrogen is produced when H2 molecules at a high temperature are irradiated with UV radiation.
   (2) At around 2000 K, the dissociation of dihydrogen into its atoms is nearly 8.1%.
   (3) Bond dissociation enthalpy of H2 is highest among diatomic gaseous molecules which contain a single bond.
   (4) Dihydrogen is produced on reacting zinc with HCl as well as NaOH(aq).

Official Ans. by NTA (2)

Sol. Atomic hydrogen is produced at high temperature in an electric arc or under ultraviolet radiations. The dissociation of dihydrogen at 2000 K is only 0.081%. H–H bond dissociation enthalpy is highest for a single bond for any diatomic molecule. Dihydrogen can be produced on reacting Zn with dil. HCl as well as NaOH (aq.)

1. Which among the following is not a polyester ?
   (1) Novolac
   (2) PHBV
   (3) Dacron
   (4) Glyptal

Official Ans. by NTA (1)

Sol. Novalac is a linear polymer of [Ph–OH + HCHO]. So ester linkage not present. So novalac is not a polyester.

1. Which one of the following correctly represents the order of stability of oxides, X2O (X = halogen)?
   (1) Br > Cl > I
   (2) Br > I > Cl
   (3) Cl > I > Br
   (4) I > Cl > Br

Official Ans. by NTA (4)

Sol. Stability of oxides of Halogens is I > Cl > Br

1. Match List-I with List-II :

List-I List-II
(Metal Ion) (Group in Qualitative analysis)
(a) Mn2+ (i) Group - III
(b) As3+ (ii) Group - IIA
(c) Cu2+ (iii) Group - IV
(d) Al3+ (iv) Group - IIB

Choose the most appropriate answer from the options given below :
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
(3) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
(4) (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Official Ans. by NTA (2)

Sol. Mn2+ → III group
As3+ → II B group
Cu2+ → II A group
Al3+ → IV group

1. The major product of the following reaction is :

(1)
(2)
(3)
(4)

Official Ans. by NTA (3)
Allen Ans. (4)

Sol. NaOH + EtOH is known as alcoholic NaOH, so it give E2 reaction with given alkyl halide.

1. For the following :

(1)
(2)
(3)
(4)

Official Ans. by NTA (2)

1. Identify correct A, B and C in the reaction sequence given below :

(1) A = nitrobenzene, B = m-chloronitrobenzene, C = m-chloroaniline
(2) A = nitrobenzene, B = m-chloronitrobenzene, C = m-chlorophenol
(3) A = nitrobenzene, B = o-chloronitrobenzene, C = o-chloroaniline
(4) A = nitrobenzene, B = p-chloronitrobenzene, C = p-chloroaniline

1. The number of S=O bonds present in sulphurous acid, peroxodisulphuric acid and pyrosulphuric acid, respectively are :
   (1) 2, 3 and 4
   (2) 1, 4 and 3
   (3) 2, 4 and 3
   (4) 1, 4 and 4

Official Ans. by NTA (4)
Sol.

SECTION-B

1. CH4 is adsorbed on 1 g charcoal at 0°C following the Freundlich adsorption isotherm. 10.0 mL of CH4 is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH4 adsorbed at 300 mm of Hg is 10^x mL. The value of x is ×10^-2.
   (Nearest integer) [Use log10 2 = 0.3010, log10 3 = 0.4771]

Official Ans. by NTA (128)

Sol. We know
x/m = K P^(1/n); using (x ∝ V)
⇒ 10/1 = K × (100)^(1/n)
15/1 = K × (200)^(1/n)
V/1 = K × (300)^(1/n)

Divide (2)/(1)
15/10 = 2^(1/n)
log(3/2) = (1/n) log 2
1/n = (log 3 – log 2)/log 2 = (0.4771 – 0.3010)/0.3010
1/n = 0.585

Divide (3)/(1)
V/10 = 3^(1/n)
log(V/10) = (1/n) log 3
log(V/10) = 0.585 × 0.4771 = 0.2791
V/10 = 10^0.279 ⇒ V = 10 × 10^0.279
⇒ V = 10^1.279 = 10^x
⇒ x = 1.279
⇒ x = 128 × 10^-2 (Nearest integer)

1. 1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol–1) and 100 g of acetone (Kb = 1.7 K kg mol–1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17°C.

2. The increase in boiling point of solution in benzene in °C is x × 10^-2. The value of x is (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O = 16.0]

Official Ans. by NTA (13)

Sol. With benzene as solvent
ΔTb = i Kb m
ΔTb = (1/2) × 2.6 × (1.22/Mw)/(100/1000)
With Acetone as solvent
ΔTb = i Kb m
0.17 = 1 × 1.7 × (1.22/Mw)/(100/1000)
(1)/(2)
ΔTb/0.17 = [(1/2)×2.6×(1.22/Mw)/(100/1000)] / [1×1.7×(1.22/Mw)/(100/1000)]
ΔTb = 0.26/2
ΔTb = 13 × 10^-2
⇒ x = 13

1. The value of magnetic quantum number of the outermost electron of Zn+ ion is

Official Ans. by NTA (0)

Sol. Zn+ → 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Outermost electron is in 4s subshell
m = 0

1. The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in AxB. The value of x is (Integer answer)

Official Ans. by NTA (1)

Sol. Anions from CCP or FCC (A–) = 4A– per unit cell
Cations occupy all octahedral voids (B–) = 4B– per unit cell
cell formula → AxBx
Empirical formula → AB
→ (x = 1)

1. In the electrolytic refining of blister copper, the total number of main impurities, from the following, removed as anode mud is
   Pb, Sb, Se, Te, Ru, Ag, Au and Pt

Official Ans. by NTA (6)

Sol. Anode mud contains Sb, Se, Te, Ag, Au and Pt

1. The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x × 10^-4. The value of x is (Nearest integer) [log 2.5 = 0.3979]

Official Ans. by NTA (6021)

Sol. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(?)
50 mL 1M, 30 mL 1M
t = 0: 50 mm, 30 mm
t = ∞: 20 mm
[HCl] = 20/80 = 1/4 M = 2.5 × 10^-1 M
pH = –log 2.5 × 10^-1 = 1 – 0.3979 = 0.6021
pH = 6021 × 10^-4

1. For the reaction A → B, the rate constant k (in s–1) is given by
   log10 k = 20.35 – (2.47×10^3)/T
   The energy of activation in kJ mol–1 is (Nearest integer)
   [Given : R = 8.314 J K–1 mol–1]

Official Ans. by NTA (47)

Sol. Given log K = 20.35 – (2.47×10^3)/T
We know log K = log A – Ea/(2.303RT)
⇒ Ea/(2.303RT) = 2.47×10^3
Ea = 2.47×10^3 × 2.303 × 8.314 kJ/mol /1000
= 47.29 = 47 (Nearest integer)

1. Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is ×10^-1 M. (Nearest integer)
   [Atomic mass : Na = 23.0, O = 16.0, H = 1.0]

Official Ans. by NTA (13)

Sol. Na2O + H2O → 2NaOH
20/62 moles
Moles of NaOH formed = 20/62 × 2

[NaOH] = (40/62)/(500/1000) = 1.29 M = 13 × 10^-1 M
(Nearest integer)

1. According to molecular orbital theory, the number of unpaired electron(s) in O2^2– is :

Official Ans. by NTA (0)

Sol. Molecular orbital configuration of O2^2– is
σ1s2 σ1s2 σ2s2 σ2s2 (π2px2 = π2py2) (π2px2 = π2py2)
Zero unpaired electron

1. The transformation occurring in Duma's method is given below :
   C2H7N + (2x + y/2) CuO → xCO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu
   The value of y is ______. (Integer answer)

Official Ans. by NTA (7)

Sol. C2H7N + (2x + y/2) CuO → xCO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu

On balancing
C2H7N + 15/2 CuO → 2CO2 + 7/2 H2O + 1/2 N2 + 15/2 Cu

On comparing
y = 7

FINAL JEE-MAIN EXAMINATION – AUGUST, 2021
(Held On Tuesday 31st August, 2021) TIME : 3 : 00 PM to 6 : 00 PM

MATHEMATICS
SECTION-A

Sol. Suppose r = xa + yb + 2c
and |a| = |b| = |c| = k
a × ((r – b) × a) + b × ((r – c) × b) + c × ((r – a) × c) = 0
⇒ k^2(r – b) – k^2xa + k^2(r – c) – k^2yb + k^2(r – a) – k^2zc = 0
⇒ 3r – (a + b + c) – r = 0
⇒ r = (a + b + c)/2

1. The domain of the function
   f(x) = sin^–1((3x^2 + x – 1)/(x – 1)^2) + cos^–1((x – 1)/(x + 1)) is :
   (1) [0, 1/4]
   (2) [–2, 0] ∪ [1/4, 1/2]
   (3) [1/4, 1/2] ∪ {0}
   (4) [0, 1/2]

Official Ans. by NTA (3)

Sol. f(x) = sin^–1((3x^2 + x – 1)/(x – 1)^2) + cos^–1((x – 1)/(x + 1))
–1 ≤ (x – 1)/(x + 1) ≤ 1 ⇒ 0 ≤ x < ∞ …(1)
–1 ≤ (3x^2 + x – 1)/(x – 1)^2 ≤ 1 ⇒ x ∈ [–1/4, 1/2] ∪ {0} …(2)
(1) & (2)
⇒ Domain = [1/4, 1/2] ∪ {0}

1. Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is :
   (1) 1/10
   (2) 1/15
   (3) 1/5
   (4) 1/30

Official Ans. by NTA (1)

1. g(3) = 2g(1) can be defined in 3 ways number of onto functions in this condition = 3 × 4!
   Total number of onto functions = 6!
   Required probability = (3 × 4!)/6! = 1/10

2. Let f : N → N be a function such that
   f(m + n) = f(m) + f(n) for every m, n ∈ N
   If f(6) = 18 then f(2)·f(3) is equal to :
   (1) 6
   (2) 54
   (3) 18
   (4) 36

Official Ans. by NTA (2)

Sol. f(m + n) = f(m) + f(n)
Put m = 1, n = 1
f(2) = 2f(1)
Put m = 2, n = 1
f(3) = f(2) + f(1) = 3f(1)
Put m = 3, n = 3
f(6) = 2f(3) ⇒ f(3) = 9
⇒ f(1) = 3, f(2) = 6
f(2)·f(3) = 6 × 9 = 54

1. The distance of the point (–1, 2, –2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x – 2y + z = 0 is :
   (1) 1/√2
   (2) 5/2
   (3) √42/2
   (4) √34/2

Official Ans. by NTA (4)

Sol. P1 : 2x + 3y + 2z = 0
⇒ n1 = 2i + 3j + 2k
P2 : x – 2y + z = 0
⇒ n2 = i – 2j + k
Direction vector of line L which is line of intersection of P1 and P2
r = n1 × n2 = 7i – 7k
DR's of L are (1, 0, –1)
⇒ Equation of L : x/1 = y/0 = z/–1 = λ

DR's of PQ = (λ + 1, –2, 2 – λ)
∴ PQ ? r
⇒ (λ + 1)(1) + (–2)(0) + (2 – λ)(–1) = 0
⇒ λ = 1/2 ⇒ Q(1/2, 0, –1/2)
⇒ PQ = √34/2

1. Negation of the statement (p ∨ r) ⇒ (q ∨ r) is :
   (1) p ∧ ¬q ∧ ¬r
   (2) ~p ∧ q ∧ ¬r
   (3) ~p ∧ q ∧ r
   (4) p ∧ q ∧ r

Official Ans. by NTA (1)

Sol. ? ~(A ⇒ B) = A ∧ ~B
∴ ~((p ∨ r) ⇒ (q ∨ r))
= (p ∨ r) ∧ (~q ∧ ~r)
= (p ∨ r) ∧ (~r) ∧ (~q)
= p ∧ (~r) ∧ (~q)

1. If α = lim_{x→π/4} (tan^3 x – tan x)/cos(x + π/4) and β = lim_{x→0} (cos x)^{cot x} are the roots of the equation, ax^2 + bx – 4 = 0, then the ordered pair (a, b) is :
   (1) (1, –3)
   (2) (–1, 3)
   (3) (–1, –3)
   (4) (1, 3)

Official Ans. by NTA (4)

Sol. α = lim_{x→π/4} (tan^3 x – tan x)/cos(x + π/4); 0/0 form
Using L Hopital rule
α = lim_{x→π/4} (3tan^2 x sec^2 x – sec^2 x)/(–sin(x + π/4))
⇒ α = –4

1. The locus of mid-points of the line segments joining (–3, –5) and the points on the ellipse x^2/4 + y^2/9 = 1 is:
   (1) 9x^2 + 4y^2 + 18x + 8y + 145 = 0
   (2) 36x^2 + 16y^2 + 90x + 56y + 145 = 0
   (3) 36x^2 + 16y^2 + 108x + 80y + 145 = 0
   (4) 36x^2 + 16y^2 + 72x + 32y + 145 = 0

Official Ans. by NTA (3)

Sol. General point on x^2/4 + y^2/9 = 1 is A(2cosθ, 3sinθ) given B(–3, –5)
midpoint C((2cosθ – 3)/2, (3sinθ – 5)/2)
h = (2cosθ – 3)/2; k = (3sinθ – 5)/2
⇒ ((2h + 3)/2)^2 + ((2k + 5)/3)^2 = 1
⇒ 36x^2 + 16y^2 + 108x + 80y + 145 = 0

1. If dy/dx = (2^2y + 2^2x^2)/(2^2 + 2^{2 + y} log2^2), y(0) = 0, then for y = 1, the value of x lies in the interval:
   (1) (1, 2)
   (2) (1/2, 1]
   (3) (2, 3)
   (4) (0, 1/2]

Official Ans. by NTA (1)

Sol. dy/dx = 2^2(y + 2^2) / (2^2(1 + 2^2 ln2))
⇒ ∫ (1 + 2^2)ln2/(y + 2^2) dy = ∫ dx
⇒ ln|y + 2^2| = x + c
x = 0; y = 0 ⇒ c = 0
⇒ x = ln|y + 2^2|
⇒ at y = 1, x = ln 3
? 3 ∈ (e, e^2) ⇒ x ∈ (1, 2)

1. An angle of intersection of the curves, x^2/a^2 + y^2/b^2 = 1 and x^2 + y^2 = ab, a > b is:
   (1) tan^–1((a + b)/√ab)
   (2) tan^–1((a – b)/2√ab)
   (3) tan^–1((a – b)/√ab)
   (4) tan^–1(2√ab)

Official Ans. by NTA (3)

Sol. x^2/a^2 + y^2/b^2 = 1, x^2 + y^2 = ab
2x1/a^2 + 2y1y'/b^2 = 0
⇒ y1 = –x1 b^2/(a^2 y1)
∴ 2x1 + 2y1y' = 0
⇒ y2 = –x1/y1
∴ (1)
Here (x1, y1) is point of intersection of both curves
∴ x1^2 = a^2b/(a + b), y1^2 = ab^2/(a + b)
∴ tan θ = |(y1 – y2)/(1 + y1y2)| = |(–x1 b^2/(a^2 y1) + x1/y1)|

1. If dy/dx = x[y^2/x^2 + φ(y^2/x^2)/φ'(y^2/x^2)], x > 0, φ > 0, and y(1) = –1, then φ(y^2/x^2) is equal to :
   (1) 4φ(2)
   (2) 4φ(1)
   (3) 2φ(1)
   (4) φ(1)

Official Ans. by NTA (2)

Sol. Let, y = tx
dy/dx = t + x dt/dx
∴ tx(t + x dt/dx) = x(t^2 + φ(t^2)/φ'(t^2))
t^2 + xt dt/dx = t^2 + φ(t^2)/φ'(t^2)
∫ t φ'(t^2)/φ(t^2) dt = ∫ dx/x …(1)

1. The sum of the roots of the equation x + 1 – 2log2(3 + 2^x) + 2log4(10 – 2^–x) = 0, is :
   (1) log2 14
   (2) log2 11
   (3) log2 12
   (4) log2 13

Official Ans. by NTA (2)

Sol. x + 1 – 2log2(3 + 2^x) + 2log4(10 – 2^–x) = 0
log2(2^{x+1}) – log2(3 + 2^x)^2 + log2(10 – 2^–x) = 0
log2(2^{x+1}(10 – 2^–x)/(3 + 2^x)^2) = 0
(2(10·2^x – 1))/(3 + 2^x)^2 = 1
⇒ 20·2^x – 2 = 9 + 2^x + 6·2^x
∴ (2^x)^2 – 14(2^x) + 11 = 0
Roots are 2^{x1} & 2^{x2}
∴ 2^{x1}·2^{x2} = 11
x1 + x2 = log2(11)

1. If z is a complex number such that (z – i)/(z – 1) is purely imaginary, then the minimum value of |z – (3 + 3i)| is :
   (1) 2√2 – 1
   (2) 3√2
   (3) 6√2
   (4) 2√2

Official Ans. by NTA (4)

Sol. (z – i)/(z – 1) is purely Imaginary number
Let z = x + iy
∴ (x + i(y – 1))/((x – 1) + iy) × ((x – 1) – iy)/((x – 1) – iy)
⇒ (x(x – 1) + y(y – 1) + i(–y – x + 1))/((x – 1)^2 + y^2) is purely
⇒ (x – 1/2)^2 + (y – 1/2)^2 = 1/2

1. Let a1, a2, a3, ... be an A.P. If (a1 + a2 + ... + a10)/(a1 + a2 + ... + ap) = 100/p^2, p ≠ 10, then a11/a10 is equal to :
   (1) 19/21
   (2) 100/121
   (3) 21/19
   (4) 121/100

Official Ans. by NTA (3)

Sol. (10/2(2a1 + 9d))/(p/2(2a1 + (p – 1)d)) = 100/p^2
(2a1 + 9d)p = 10(2a1 + (p – 1)d)
9dp = 20a1 – 2pa1 + 10d(p – 1)
9p = (20 – 2p)a1/d + 10(p – 1)
a1/d = (10 – p)/(2(10 – p)) = 1/2
∴ a11/a10 = (a1 + 10d)/(a1 + 9d) = (1/2 + 10)/(1/2 + 9) = 21/19

1. Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :
   (1) 4/√5
   (2) 16/√5
   (3) 8/√5
   (4) 12/√5

Official Ans. by NTA (3)

Sol.
4α – 2β = ±24 + 8
⇒ 4α – 2β = +24 + 8 ⇒ 2α – β = 16
2x – y – 16 = 0 …(1)
⇒ 4α – 2β = –24 + 8 ⇒ 2α – β = –8
2x – y + 8 = 0 …(2)
perpendicular distance of (1) from (0, 0)
|0 – 0 – 16|/√5 = 16/√5
perpendicular distance of (2) from (0, 0) is
|0 – 0 + 8|/√5 = 8/√5

1. The number of solutions of the equation
   32^{tan^2 x} + 32^{sec^2 x} = 81, 0 ≤ x ≤ π/4 is :
   (1) 3
   (2) 1
   (3) 0
   (4) 2

Official Ans. by NTA (2)

Sol. (32)^{tan^2 x} + (32)^{sec^2 x} = 81
⇒ (32)^{tan^2 x} + (32)^{1 + tan^2 x} = 81
⇒ (32)^{tan^2 x} = 81/33
In interval [0, π/4] only one solution

1. Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
   (1) f''(x) = 0 for all x ∈ (0, 2)
   (2) f''(x) = 0 for some x ∈ (0, 2)
   (3) f'(x) = 0 for some x ∈ [0, 2]
   (4) f''(x) > 0 for all x ∈ (0, 2)

Official Ans. by NTA (2)

Sol. f(0) = 0, f(1) = 1 and f(2) = 2
Let h(x) = f(x) – x has three roots

1. If [x] is the greatest integer ≤ x, then
   π^2 ∫_0^2 (sin(πx/2))(x – [x])^{[x]} dx is equal to :
   (1) 2(π – 1)
   (2) 4(π – 1)
   (3) 4(π + 1)
   (4) 2(π + 1)

Official Ans. by NTA (2)

π^2 [∫_0^1 sin(πx/2) dx + ∫_1^2 sin(πx/2)(x – 1) dx]
= π^2 [–(2/π)(cos(πx/2)) + (x – 1)(–(2/π)cos(πx/2))|_1^2 – (2/π)cos(πx/2) dx]
= π^2 [0 + 2/π + 2/π + 2/π + (2/π)(sin(πx/2))^2]
= 4π – 4 = 4(π – 1)

1. The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :
   (1) 92/5
   (2) 134/5
   (3) 536/25
   (4) 112/5

Official Ans. by NTA (3)

Sol. Let 8, 16, x1, x2, x3, x4, x5 be the observations.
Now (x1 + x2 + ... + x5 + 14)/7 = 8
⇒ ∑{i=1}^5 x_i = 42 …(1)
Also (x1^2 + x2^2 + ... x5^2 + 8^2 + 6^2)/7 – 64 = 16
⇒ ∑{i=1}^5 x_i^2 = 560 – 100 = 460 …(2)
So variance of x1, x2, ..., x5
= 460/5 – (42/5)^2 = (2300 – 1764)/25 = 536/25

SECTION-B

1. If the coefficient of a^2 b^8 in the expansion of (a + 2b + 4ab)^10 is K·2^16, then K is equal to

Official Ans. by NTA (315)

10!/(α! β! γ!) a^α (2b)^β (4ab)^γ
= 10!/(α! β! γ!) a^{α+γ} b^{β+γ} 2^β 4^γ
α + β + γ = 10 …(1)
α + γ = 7 …(2)
β + γ = 8 …(3)
(2) + (3) – (1) ⇒ γ = 5
α = 2
β = 3
so coefficients = 10!/(2!3!5!) 2^3 2^10
= (10×9×8×7×6×5)/(2×3×2×5!) × 2^13
= 315 × 2^16 ⇒ k = 315

1. Suppose the line (x – 2)/α = (y – 2)/–5 = (z + 2)/2 lies on the plane x + 3y – 2z + β = 0. Then (α + β) is equal to

Official Ans. by NTA (7)

Sol. Point (2, 2, –2) also lies on given plane
So 2 + 3×2 – 2(–2) + β = 0
⇒ 2 + 6 + 4 + β = 0 ⇒ β = –12
Also α×1 – 5×3 + 2×–2 = 0
⇒ α – 15 – 4 = 0 ⇒ α = 19
∴ α + β = 19 – 12 = 7

1. The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is

Official Ans. by NTA (5143)

Sol. A = 4-digit numbers divisible by 3
A = 1002, 1005, ..., 9999
9999 = 1002 + (n – 1)3
⇒ (n – 1)3 = 8997 ⇒ n = 3000
B = 4-digit numbers divisible by 7
B = 1001, 1008, ..., 9996
⇒ 9996 = 1001 + (n – 1)7
⇒ n = 1286
A ∩ B = 1008, 1029, ..., 9996
9996 = 1008 + (n – 1)21
⇒ n = 429
So, no divisible by either 3 or 7
= 3000 + 1286 – 429 = 3857
total 4-digits numbers = 9000
required numbers = 9000 – 3857 = 5143

1. If ∫ sin x/(sin^3 x + cos^3 x) dx =
   α log_e |1 + tan x| + β log_e |1 – tan x + tan^2 x| + γ tan^–1((2tan x – 1)/√3) + C,
   when C is constant of integration, then the value of 18(α + β + γ^2) is

Official Ans. by NTA 3

Sol. = ∫ sin x/(1 + tan^3 x) dx = ∫ tan x·sec^2 x/((tan x + 1)(1 + tan^2 x – tan x)) dx
Let tan x = t ⇒ sec^2 x dx = dt
= ∫ t/((t + 1)(t^2 – t + 1)) dt
= ∫ (A/(t + 1) + B(2t – 1)/(t^2 – t + 1) + C/(t^2 – t + 1)) dx
= A(t^2 – t + 1) + B(2t – 1)(t^2 – t + 1) + C(t + 1) = t
= t^2(A + 2B) + t(–A + B + C) + A – B + C = 1
∴ A + 2B = 0 …(1)
–A + B + C = 1 …(2)
A – B + C = 0 …(3)
⇒ C = 1/2 ⇒ A – B = –1/2 …(4)
A + 2B = 0
A – B = –1/2
⇒ 3B = 1/2 ⇒ B = 1/6
A = –1/3
I = –1/3 ∫ dt/(1 + t) + 1/6 ∫ (2t – 1)/(t^2 – t + 1) dt + 1/2 ∫ dt/(t^2 – t + 1)
= –1/3 ln|1 + tan x| + 1/6 ln|tan^2 x – tan x + 1|

• 1/2·2/√3 tan^–1[((tan x – 1/2)/√3)]
  = –1/3 ln|1 + tan x| + 1/6 ln|tan^2 x – tan x + 1|

• 1/√3 tan^–1((2tan x – 1)/√3) + C
  α = –1/3, β = 1/6, γ = 1/√3
  18(α + β + γ^2) = 18(–1/3 + 1/6 + 1/3) = 3

1. A tangent line L is drawn at the point (2, –4) on the parabola y^2 = 8x. If the line L is also tangent to the circle x^2 + y^2 = a, then 'a' is equal to

Official Ans. by NTA 2

Sol. tangent of y^2 = 8x is y = mx + 2/m
P(2, –4) ⇒ –4 = 2m + 2/m
⇒ m + 1/m = –2 ⇒ m = –1
∴ tangent is y = –x – 2
⇒ x + y + 2 = 0 …(1)

(1) is also tangent to x^2 + y^2 = a
So 2/√2 = √a ⇒ √a = √2
⇒ a = 2

1. If S = 7/5 + 9/5^2 + 13/5^3 + 19/5^4 + ... then 160 S is equal to

Official Ans. by NTA (305)

Sol. S = 7/5 + 9/5^2 + 13/5^3 + 19/5^4 + ...
(1/5)S = 7/5^2 + 9/5^3 + 13/5^4 + ...
On subtracting
(4/5)S = 7/5 + 2/5^2 + 4/5^3 + 6/5^4 + ...
S = 7/4 + 1/10(1 + 2/5 + 3/5^2 + ...)
S = 7/4 + 1/10(1 – 1/5)^–2
= 7/4 + 1/10 × 25/16 = 61/32
⇒ 160S = 5 × 61 = 305

1. The number of elements in the set
   {A = [a b; 0 d]; a, b, d ∈ {–1, 0, 1} and (I – A)^3 = I – A^3},
   where I is 2 × 2 identity matrix, is :

Official Ans. by NTA (8)

(I – A)^3 = I^3 – A^3 – 3A(I – A) = I – A^3
⇒ 3A(I – A) = 0 or A^2 = A
⇒ a^2 = a, b(a + d – 1) = 0, d^2 = d
If b ≠ 0, a + d = 1 ⇒ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 ⇒ 4 ways
⇒ Total 8 matrices

1. If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x – x^2, then 12m is equal to

Official Ans. by NTA (26)

Sol. Total area = ∫_0^{3/2} (1 + 4x – x^2) dx
= x + 2x^2 – x^3/3 |_0^{3/2} = 39/8
& 39/16 = 1/2 · 3/2 · 3/2 m
⇒ 3m = 13/2 ⇒ 12m = 26

1. Let B be the centre of the circle x^2 + y^2 – 2x + 4y + 1 = 0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8·(area ΔAPQ / area ΔBPQ) is equal to

Official Ans. by NTA (18)

Sol.

1. Let f(x) be a cubic polynomial with f(1) = –10, f(–1) = 6 and has a local minima at x = 1 and f'(x) has a local minima at x = –1. Then f(3) is equal to

Official Ans. by NTA (22)

Sol. F'(x) = a(x – 1)(x + 3)
F''(x) = 6a(x + 1)
F'(x) = 3a(x + 1)^2 + b
F'(1) = 0 ⇒ b = –12a
F(x) = a(x + 1)^3 – 12ax + c
= (x + 1)^3 – 12x – 6
F(3) = 64 – 36 – 6 = 22

1. A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
   (1) √((2ghv^2 + 1)/h^2)
   (2) √((2v^2h)/g + h^2)
   (3) √((2gh)/v^2 + h^2)
   (4) √((2gh)/v^2 + h^2)

Official Ans. by NTA (3)

R = √(2h/g)·v
D = √(R^2 + h^2)
= √((2h/g·v)^2 + h^2)
D = √(2hv^2/g + h^2)
Option (3) is correct

1. In the following logic circuit the sequence of the inputs A, B are (0, 0), (0,1), (1, 0) and (1, 1). The output Y for this sequence will be :
   (1) 1,0,1,0
   (2) 0,1,0,1
   (3) 1,1,1,0
   (4) 0,0,1,1

Official Ans. by NTA (3)

Sol. Y = (A·B)·(A + B)
Y|(0,0) = 1
Y|(0,1) = 1
Y|(1,0) = 1
Y|(1,1) = 0
Option (3) is correct

1. Two particles A and B having charges 20μC and –5μC respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?
   20μC A 5cm B –5μC
   (1) At 5 cm from 20μC on the left side of system
   (2) At 5 cm from –5μC on the right side
   (3) At 1.25 cm from –5μC between two charges
   (4) At midpoint between two charges

Official Ans. by NTA (2)

Sol. 20μC –5μC
Null point is possible only right side of –5μC
E_N = + k(–5μC)/x^2 + k(20μC)/(5 + x)^2 = 0
x = 5 cm
∴ option (2) is correct

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