Official NTA Question PaperFull Solved Paper2022Evening Shift
JEE Main 2022 28th July Evening Shift Question Paper with Solutions
Official solved question paper for Joint Entrance Examination (Main). Review all 90 questions, official answer keys, and detailed step-by-step solutions below, or attempt online with timer and negative marking.
A uniform electric field E=(8m/e)V/m is created between two parallel plates of length 1m as shown in figure, (where m= mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2m/s. The angle of the deviation (θ) of the path of the electron as it comes out of the field will be _________.
A
tan−1(4)
B
tan−1(2)
C
tan−1(31)
D
tan−1(3)
Official Correct Answer:Option (B)
Step-by-Step Solution:
E=e8m V/m
l=1 m
vx=2 m/s
ay=−8 m/s2
t=vxl=21s
⇒∣vy∣=4 m/s
⇒ angle of deviation = θ
tanθ=vxvy
θ=tan−1(24)=tan−1(2)
Q 2PhysicsSection B • Numericalelectromagnetic induction
+4 / -1
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _________ A.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
Just after closing the switch, i=4+26=1A
Q 3PhysicsSection A • MCQcapacitor
+4 / -1
A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 43d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given C0 = capacitance of capacitor with air as medium between plates.)
A
3+K4KC0
B
3+K3KC0
C
4KC03+K
D
4+KK
Official Correct Answer:Option (A)
Step-by-Step Solution:
C0=dε0A
C=d−43d+4K3dε0A=3d+Kd4ε0AK
=3+K4KC0
Q 4PhysicsSection A • MCQmagnetics
+4 / -1
The magnetic field at the center of current carrying circular loop is B1. The magnetic field at a distance of 3 times radius of the given circular loop from the center on its axis is B2. The value of B1/B2 will be
A
9 : 4
B
12 : 5
C
8 : 1
D
5 : 3
Official Correct Answer:Option (C)
Step-by-Step Solution:
B1=2Rμ0i
B2=2(R2+x2)23μ0iR2
⇒B2B1=R31(R2+x2)23
=R31(8R3)
=8
Q 5PhysicsSection B • Numericalmotion
+4 / -1
A ball is thrown vertically upwards with a velocity of 19.6ms−1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be (5k)m. The value of k is __________. (use g=9.8m/s2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 392
Official Correct Answer:392
Step-by-Step Solution:
v = 19.6 m/s
t = 6s
Time taken in upward motion above tower = 2s
⇒ Time taken from top most point to ground = 4s
⇒g2h=4
h=216×9.8=8×9.8
⇒k=8×9.8×5=392
Q 6PhysicsSection A • MCQray optics
+4 / -1
The power of a lens (biconvex) is 1.25m−1 in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20cm and 40cm respectively. The refractive index of surrounding medium:
A
1.0
B
79
C
23
D
34
Official Correct Answer:Option (B)
Step-by-Step Solution:
∵f1=(μ1μ2−1)(R11−R21)
⇒1001.25=(μ11.5−1)(201+401)
⇒801=(μ11.5−1)×80(4+2)
⇒μ11.5−1=61
⇒μ11.5=67
⇒μ1=71.5×6=79
Q 7PhysicsSection A • MCQmotion
+4 / -1
At time t=0 a particle starts travelling from a height 7z^cm in a plane keeping z coordinate constant. At any instant of time it's position along the x^ and y^ directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be
A
−30y^
B
30y^
C
3x^+15y^
D
3x^+15y^+7z^
Official Correct Answer:Option (B)
Step-by-Step Solution:
x=3t⇒ax=0
y=5t3⇒ay=30t
a(t=1)=30y
Q 8PhysicsSection B • Numericalheat and thermodynamics
+4 / -1
At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be _________ J.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 750
Official Correct Answer:750
Step-by-Step Solution:
f=8
W=PdV=150
Q=W+ΔU
=PdV+2fPdV
Q=5×150=750J
Q 9PhysicsSection B • Numericalcurrent electricity
+4 / -1
An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 14
This power developed is proportional to resistance.
So, P60′=Pnet×16060=37.5×16060≃14W
Q 10PhysicsSection A • MCQelectronic devices
+4 / -1
An n.p.n transistor with current gain β=100 in common emitter configuration is shown in figure. The output voltage of the amplifier will be
A
0.1 V
B
1.0 V
C
10 V
D
100 V
Official Correct Answer:Option (B)
Step-by-Step Solution:
ViV0=β×(RBRC)
⇒V0=100×(110)×10−3
=1.0 V
Q 11PhysicsSection B • Numericalray optics
+4 / -1
An object 'O' is placed at a distance of 100cm in front of a concave mirror of radius of curvature 200cm as shown in the figure. The object starts moving towards the mirror at a speed 2cm/s. The position of the image from the mirror after 10s will be at _________ cm.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 400
Official Correct Answer:400
Step-by-Step Solution:
The object after 10 second will be at u=−80 cm.
So v1−801=−1001
⇒v=+208000=400 cm
Q 12PhysicsSection A • MCQwork power and energy
+4 / -1
A bullet of mass 200g having initial kinetic energy 90J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40J within 1s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is
A
45 m
B
90 m
C
125 m
D
25 m
Official Correct Answer:Option (A)
Step-by-Step Solution:
21mx2=90
⇒21×0.2×x2=90,
x2=900
x=30 m/s
21mv2=40⇒v=32×30=20 m/s
20=30−a×1⇒a=−10 m/s2
0−x2=2as
s=−2ax2=2×1030×30
=45 m
Q 13PhysicsSection A • MCQcurrent electricity
+4 / -1
Given below are two statements :
Statement I : A uniform wire of resistance 80Ω is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5Ω.
Statement II: Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3:2.
In the light of the above statements, choose the most appropriate answer from the option given below
A
Both statement I and statement II are correct
B
Both statement I and statement II are incorrect
C
Statement I is correct but statement II is incorrect
D
Statement I is incorrect but statement II is correct
Official Correct Answer:Option (C)
Step-by-Step Solution:
Statement I : R1part=480=20Ω
⇒Reff=420=5Ω
Statement II : Ratio =2R(ΔV)23R(ΔV)2
=32
Q 14PhysicsSection A • MCQalternating current
+4 / -1
A transformer operating at primary voltage 8kV and secondary voltage 160V serves a load of 80kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
A
800Ω and 1.06Ω
B
10Ω and 500Ω
C
800Ω and 0.32Ω
D
1.06Ω and 500Ω
Official Correct Answer:Option (C)
Step-by-Step Solution:
V1i1=V2i2=80 kW
⇒i1=10A and i2=16080×1000=500A
⇒R1=i1V1=800Ω and R2=500160=0.32Ω
Q 15PhysicsSection A • MCQmagnetics
+4 / -1
A triangular shaped wire carrying 10A current is placed in a uniform magnetic field of 0.5T, as shown in figure. The magnetic force on segment CD is
(Given BC=CD=BD=5cm.)
A
0.126 N
B
0.312 N
C
0.216 N
D
0.245 N
Official Correct Answer:Option (C)
Step-by-Step Solution:
F=il×B
=ilBsin60∘
=10×1005×0.5×23
=0.2165 N
Q 16PhysicsSection A • MCQdual nature of radiation
+4 / -1
Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be
A
1 : 2
B
1 : 3
C
2 : 3
D
3 : 2
Official Correct Answer:Option (C)
Step-by-Step Solution:
21mv12=5ϕ−ϕ
And, 21mv22=10ϕ−ϕ
⇒(v2v1)2=94
⇒v2v1=32
Q 17PhysicsSection A • MCQelectromagnetic waves
+4 / -1
Sun light falls normally on a surface of area 36cm2 and exerts an average force of 7.2×10−9N within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is
A
25.92×102W/cm2
B
8.64×10−6W/cm2
C
6.0W/cm2
D
0.06W/cm2
Official Correct Answer:Option (D)
Step-by-Step Solution:
Pressure =cl
⇒AF=cl
⇒l=36×10−47.2×10−9×3×108 W/m2
=600 W/m2
⇒l=0.06 W/cm2
Q 18PhysicsSection A • MCQproperties of matter
+4 / -1
Consider a cylindrical tank of radius 1m is filled with water. The top surface of water is at 15m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5m from the bottom. A force of 5×105N is applied an the top surface of water using a piston. The speed of ifflux from the hole will be : (given atmospheric pressure PA=1.01×105Pa, density of water ρW=1000kg/m3 and gravitational acceleration g=10m/s2 )
A
11.6 m/s
B
10.8 m/s
C
17.8 m/s
D
14.4 m/s
Official Correct Answer:Option (C)
Step-by-Step Solution:
By Bernoulli's theorem,
π(1)25×105+ρg(10)=1.01×105+21ρ(v)2
⇒v2=200+1000π106−202
⇒v≃17.8 m/s
Q 19PhysicsSection A • MCQatoms and nuclei
+4 / -1
A radioactive sample decays 87 times its original quantity in 15 minutes. The half-life of the sample is
A
5 min
B
7.5 min
C
15 min
D
30 min
Official Correct Answer:Option (A)
Step-by-Step Solution:
N=2T1/2tN0
⇒2T1/2t=NN0=(8N0)N0=8
⇒T1/2t=3
⇒T1/2=315=5 min
Q 20PhysicsSection A • MCQlaws of motion
+4 / -1
A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length 'L−l' is hanging on the other side of the pully. At a certain point of time, when l=xL, the acceleration of the chain is 2g. The value of x is __________.
A FM Broadcast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is :
A
220 kHz
B
180 kHz
C
360 kHz
D
440 khz
Official Correct Answer:Option (D)
Step-by-Step Solution:
Bandwidth of FM wave =2(Δf+fm)
fmΔf=10 (Given)
Δf=fm(10)=20×10=200 kHz
BW=2(200+20) kHz
=440 kHz
Q 22PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A vessel contains 14g of nitrogen gas at a temperature of 27∘C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :
Take R=8.32Jmol−1k−1.
A
2229 J
B
5616 J
C
9360 J
D
13,104 J
Official Correct Answer:Option (C)
Step-by-Step Solution:
n = 0.5
T = 300
For vrms to be doubled T' = 4 × 300 = 1200
⇒ Heat transferred
=(0.5)(25)(8.32)(900)
=9360 J
Q 23PhysicsSection B • Numericalproperties of matter
+4 / -1
A string of area of cross-section 4mm2 and length 0.5m is connected with a rigid body of mass 2kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires a speed of 5m/s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _________ ×10−5.
(use Young's modulus 1011N/m2 and g=10m/s2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 30
Official Correct Answer:30
Step-by-Step Solution:
A=4×10−6 m2
l=0.5 m
m=2 kg
vb=5 m/s
Tb=mg+m(lVb2)
=20+2×2125=120 N
lΔl=ATb×Y1=4×10−6120×10−11=30×10−5
Q 24PhysicsSection B • Numericalsimple harmonic motion
+4 / -1
The potential energy of a particle of mass 4kg in motion along the x-axis is given by U=4(1−cos4x) J. The time period of the particle for small oscillation (sinθ≃θ) is (Kπ)s. The value of K is _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
U=4(1−cos4x)
⇒F=−dxdU=−(4)(4sin4x)
=−16sin4x
as small x
F=−16(4x)=−64x≡−kx
T=2πkm=2π644=2π
⇒K=2
Q 25PhysicsSection B • Numericalray optics
+4 / -1
In an experiment with a convex lens, The plot of the image distance (v′) against the object distance (μ′) measured from the focus gives a curve v′μ′=225. If all the distances are measured in cm. The magnitude of the focal length of the lens is ___________ cm.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 15
Official Correct Answer:15
Step-by-Step Solution:
Using Newton's formula for lenses,
v′μ′=f2=225⇒f=15
Q 26PhysicsSection B • Numericalcenter of mass
+4 / -1
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ=ρ0(1−L2x2)kg/m and length L (in meter) is α3Lm. The value of α is ___________. (where x is the distance from end A)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 8
Official Correct Answer:8
Step-by-Step Solution:
ρ=ρ0(1−L2x2) kg/m
xcm=A0∫Lρ0(1−L2x2)dxA0∫Lρ0(1−L2x2)xdx
xcm=L−3L2L2−4L2=32L4L2=83L
⇒α=8
Q 27PhysicsSection A • MCQsimple harmonic motion
+4 / -1
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -
(consider radius of earth RE=6400km and g on earth 10m/s2 )
A
1200 km
B
1600 km
C
3200 km
D
4800 km
Official Correct Answer:Option (C)
Step-by-Step Solution:
T∝1/g
⇒T2T1=g1g2=R+hR
64=R+hR
⇒h=R/2
=3200 km
Q 28PhysicsSection A • MCQunits and measurements
+4 / -1
Consider the efficiency of carnot's engine is given by η=sinθαβlogekTβx, where α and β are constants. If T is temperature, k is Boltzmann constant, θ is angular displacement and x has the dimensions of length. Then, choose the incorrect option :
A
Dimensions of β is same as that of force.
B
Dimensions of α−1x is same as that of energy.
C
Dimensions of η−1sinθ is same as that of αβ.
D
Dimensions of α is same as that of β.
Official Correct Answer:Option (D)
Step-by-Step Solution:
Since, dimensions trigonometric function and logarithmic function are dimensionless quantities.
∴[η]=[M0L0T0]<br><br>
Also, dimensions of temperature, [T]=[M0L0T0K]
Dimensions of Boltzmann constant, [k]=[ML2T−2K−1]
Dimension of x=[M0LT0]
(A) [β]=[xkT]=[xE]=[MLT−2]=[F]
(B) [αβ]=[M0L0T0]
[α]−1=[β]=[xkT]
So [α]−1[x]=[kT]=[ML2T−2]
(C) ηsinθ=αβ
So [ηsinθ]=[αβ]
[η]=[M0L0T0] it is dimensionless quantity
(D) [α]=[β]
Q 29PhysicsSection B • Numericalunits and measurements
+4 / -1
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5s is measured from time of 100 oscillation with a watch of 1s resolution. If measured value of length is 10cm known to 1mm accuracy, The accuracy in the determination of g is found to be x%. The value of x is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
T=2πgl
gdg×100=T2dT×100+ldl×100
=2×501×100+1001×100=5%
Q 30PhysicsSection A • MCQproperties of matter
+4 / -1
A pressure-pump has a horizontal tube of cross sectional area 10cm2 for the outflow of water at a speed of 20m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :
[given: density of water =1000kg/m3]
A
300 N
B
500 N
C
250 N
D
400 N
Official Correct Answer:Option (D)
Step-by-Step Solution:
Fw=ρAv2
=103×10×10−4×20×20
=400N
Q 31ChemistrySection B • Numericalsolid state
+4 / -1
Metal M crystallizes into a fcc lattice with the edge length of 4.0×10−8cm. The atomic mass of the metal is ____________ g/mol. (Nearest integer)
( Use : NA=6.02×1023mol−1, density of metal, M=9.03gcm−3 )
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 87
Official Correct Answer:87
Step-by-Step Solution:
ρ=NAa3ZM⇒M=49.03×6.02×1023×(4×10−8)3
=49.03×6.02×64×10−1=86.9gmol−1≈87gmol−1
Q 32ChemistrySection B • Numericalchemical kinetics and nuclear chemistry
+4 / -1
For a reaction, given below is the graph of lnk vs T1. The activation energy for the reaction is equal to ____________ calmol−1. (nearest integer)
(Given : R=2calK−1mol−1 )
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 8
Q 33ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Thin layer chromatography is an adsorption chromatography.
Reason R : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.
In the light of the above statements, choose the correct answer from the options given below
A
Both A and R are true and R is the correct explanation of A
B
Both A and R are true but R is NOT the correct explanation of A
C
A is true but R is false
D
A is false but R is true
Official Correct Answer:Option (A)
Step-by-Step Solution:
Thin layer chromoatography (TLC) is another type
of adsorption chromatography, which involve
sepration of substance of a mixture ovel a thin
layer of an adsorbent coated on glass plate.
A thin layer (about 0.2 mm thick) of an adsorbent
(silica gel) or (Alumina) in spread overa glass plate
of suitable size. Hence Assertion (A) is correct and
Reason (R) is correct explanation of (A)
Q 34ChemistrySection B • Numericalatomic structure
+4 / -1
If the wavelength for an electron emitted from H-atom is 3.3×10−10m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
[ Given :h=6.626×10−34Js ]
Mass of electron =9.1×10−31kg
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Total energy absorbed = lonization energy + Kinetic energy
=(21.76+21.8)×10−19=43.56×10−19J≈2 times of 21.76×10−19J
Q 35ChemistrySection B • Numericalchemical equilibrium
+4 / -1
At 600K,2mol of NO are mixed with 1mol of O2.
2NO(g)+O2(g)⇄2NO2(g)
The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid.
Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below
A
Both A and R are true and R is the correct explanation of A
B
Both A and R are true but R is NOT the correct explanation of A
C
A is true but R is false
D
A is false but R is true
Official Correct Answer:Option (A)
Step-by-Step Solution:
2KMnO4+16HCl→2MnCl2+2KCl+8H2O+Cl2
HCl is not used in the process of titration because it reacts with the (KMnO4) that is used in the process and gets oxidized.
Q 37ChemistrySection A • MCQthermodynamics
+4 / -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason R : The value of ΔGΘ becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements, choose the most appropriate answer from the options given below
A
Both A and R are correct and R is the correct explanation of A
B
Both A and R are correct but R is NOT the correct explanation of A
C
A is correct but R is not correct
D
A is not correct but R is correct
Official Correct Answer:Option (A)
Step-by-Step Solution:
Reduction of a metal oxide is easier if the metal is formed in a liquid state at the temperature of reduction because the entropy is higher if the metal is in a liquid state.
Q 38ChemistrySection B • Numericalhydrogen
+4 / -1
On reaction with stronger oxidizing agent like KIO4, hydrogen peroxide oxidizes with the evolution of O2. The oxidation number of I in KIO4 changes to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
KIO4+7+H2O2⟶KIO3+5+H2O+O2
Q 39ChemistrySection B • Numericalsurface chemistry
+4 / -1
Among the following the number of curves not in accordance with Freundlich adsorption isotherm is ________.
(a)
(b)
(c)
(d)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 3
Official Correct Answer:3
Step-by-Step Solution:
The following curves are not in accordance with
Freundlich adsorption isotherm.
Q 40ChemistrySection A • MCQpolymers
+4 / -1
Match List I with List II
List - I (Polymer)
List - II (Nature)
(A)
(I)
Thermosetting polymer
(B)
(II)
Fibers
(C)
(III)
Elastomer
(D)
(IV)
Thermoplastic polymer
Choose the correct answer from the options given below :
A
A-II, B-III, C-IV, D-I
B
A-III, B-II, C-IV, D-I
C
A-III, B-I, C-IV, D-II
D
A-I, B-III, C-IV, D-II
Official Correct Answer:Option (B)
Step-by-Step Solution:
Q 41ChemistrySection A • MCQhydrogen
+4 / -1
The products obtained during treatment of hard water using Clark's method are :
Species having same number of electrons are isoelectronic
Tb⟶65Tb+2⟶63 electrons Tm⟶69Tm+4⟶65 electrons } not isoelectronic
Sm+2⟶60 electrons Er+3⟶65 electrons } not isoelectronic
Q 43ChemistrySection A • MCQchemistry in everyday life
+4 / -1
Two statements in respect of drug-enzyme interaction are given below
Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme.
Statement II : An inhibitor can form a strong covalent bond with the enzyme.
In the light of the above statements, choose the correct answer from the options given below
A
Both Statement I and Statement II are true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true
Official Correct Answer:Option (D)
Step-by-Step Solution:
Let's analyze both statements:
Statement I: Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme. This statement is false because enzyme inhibition can occur in more than one way. Besides competitive inhibition (where the inhibitor binds to the active site), there are other types such as non-competitive and uncompetitive inhibition, where the inhibitor can bind elsewhere on the enzyme, altering its function.
Statement II: An inhibitor can form a strong covalent bond with the enzyme. This statement is true. Some inhibitors, known as irreversible inhibitors, form strong covalent bonds with the enzyme, permanently inactivating it.
So the correct answer is:
Option D: Statement I is false but Statement II is true.
Q 44ChemistrySection A • MCQs block elements
+4 / -1
Statement I : An alloy of lithium and magnesium is used to make aircraft plates.
Statement II : The magnesium ions are important for cell-membrane integrity.
In the light the above statements, choose the correct answer from the options given below :
A
Both Statement I and Statement II are true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true
Official Correct Answer:Option (B)
Step-by-Step Solution:
Alloy of Li and Mg is used to make armour plates
and not aircraft plates.
Calcium plays important roles in neuromuscular
function, interneuronal transmission and cell
membrane integrity .
Q 45ChemistrySection A • MCQcompounds containing nitrogen
+4 / -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Aniline on nitration yields ortho, meta & para nitro derivatives of aniline.
Reason R : Nitrating mixture is a strong acidic mixture.
In the light of the above statements, choose the correct answer from the options given below
A
Both A and R are true and R is the correct explanation of A
B
Both A and R are true but R is NOT the correct explanation of A
C
A is true but R is false
D
A is false but R is true
Official Correct Answer:Option (A)
Step-by-Step Solution:
Due to formation of anilinium ion in acidic
medium meta product is also obtained in
significant amount
Q 46ChemistrySection A • MCQp block elements
+4 / -1
Concentrated HNO3 reacts with Iodine to give :
A
HI, NO2 and H2O
B
HIO2,N2O and H2O
C
HIO3,NO2 and H2O
D
HIO4,NO2 and H2O
Official Correct Answer:Option (C)
Step-by-Step Solution:
I2+10HNO3(conc)→2HIO3+10NO2+4H2O
Q 47ChemistrySection A • MCQatomic structure
+4 / -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Zero orbital overlap is an out of phase overlap.
Reason R : It results due to different orientation / direction of approach of orbitals.
In the light of the above statements, choose the correct answer from the options given below
A
Both A and R are true and R is the correct explanation of A
B
Both A and R are true but R is NOT the correct explanation of A
C
A is true but R is false
D
A is false but R is true
Official Correct Answer:Option (A)
Step-by-Step Solution:
Zero overlapping is something in which there is no overlapping between two orbitals. The first condition is that the two orbitals should not be symmetrical and the second condition is that both orbitals should be in different planes.
Q 48ChemistrySection A • MCQbiomolecules
+4 / -1
The formulas of A and B for the following reaction sequence
are
A
A=C7H14O8,B=C6H14
B
A=C7H13O7,B=C7H14O
C
A=C7H12O8,B=C6H14
D
A=C7H14O8,B=C6H14O6
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 49ChemistrySection A • MCQcoordination compounds
+4 / -1
Match List I with List II
List - I (Complex)
List - II (Hybridization)
(A)
Ni(CO)4
(I)
sp3
(B)
[Ni(CN)4]2−
(II)
sp3d2
(C)
[Co(CN)6]3−
(III)
d2sp3
(D)
[CoF6]3−
(IV)
dsp2
Choose the correct answer from the options given below :
Q 50ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
The major product in the given reaction is
A
B
C
D
Official Correct Answer:Option (C)
Step-by-Step Solution:
Q 51ChemistrySection A • MCQs block elements
+4 / -1
The correct decreasing order for metallic character is :
A
Na > Mg > Be > Si > P
B
P > Si > Be > Mg > Na
C
Si > P > Be > Na > Mg
D
Be > Na > Mg > Si > P
Official Correct Answer:Option (A)
Step-by-Step Solution:
Metallic character increases top to bottom in group and decreases left to right in a period.
Mg is from second group it will be less metallic than Na. Be comes above Mg hence less metallic than Mg. Si is more metallic than phosphorous.
Q 52ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
2L of 0.2M H2SO4 is reacted with 2L of 0.1M NaOH solution, the molarity of the resulting product Na2SO4 in the solution is _________ millimolar. (Nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 25
Official Correct Answer:25
Step-by-Step Solution:
Molarity of Na2SO4 is 40.1=0.025M
=25mM
Q 53ChemistrySection A • MCQp block elements
+4 / -1
White phosphorus reacts with thionyl chloride to give :
A
PCl5,SO2 and S2Cl2
B
PCl3,SO2 and S2Cl2
C
PCl3,SO2 and Cl2
D
PCl5,SO2 and Cl2
Official Correct Answer:Option (B)
Step-by-Step Solution:
White phosphorous P4+ Thionyl chloride 8SOCl2⟶4PCl3+4SO2+2S2Cl2
Q 54ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
Arrange the following in increasing order of reactivity towards nitration
A. p-xylene
B. bromobenzene
C. mesitylene
D. nitrobenzene
E. benzene
Choose the correct answer from the options given below
A
C < D < E < A < B
B
D < B < E < A < C
C
D < C < E < A < B
D
C < D < E < B < A
Official Correct Answer:Option (B)
Step-by-Step Solution:
The correct order of reactivity towards nitration is
as electron releasing groups on benzene ring facilitate the nitration at benzene ring.
Q 55ChemistrySection B • Numericalpractical organic chemistry
+4 / -1
A sample of 0.125g of an organic compound when analyzed by Duma's method yields 22.78mL of nitrogen gas collected over KOH solution at 280K and 759mmHg. The percentage of nitrogen in the given organic compound is __________. (Nearest integer)
Given :
(a) The vapour pressure of water of 280K is 14.2mmHg.
(b) R=0.082L atm K−1mol−1
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 22
A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is __________ atm. (Nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
Given that XA=0.2,YA=0.5,PT=0.8atm
We know that PA=YA×PT
PA=0.5×0.8=0.4
Now PA=XA×PA∘⇒PA∘=0.20.4=2atm
Q 57ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
Find out the major product for the above reaction.
A
B
C
D
Official Correct Answer:Option (C)
Step-by-Step Solution:
Q 58ChemistrySection A • MCQalcohols phenols and ethers
+4 / -1
Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Q 59ChemistrySection B • Numericalthermodynamics
+4 / -1
Among the following the number of state variables is ______________.
Internal energy (U)
Volume (V)
Heat (q)
Enthalpy (H)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 3
Official Correct Answer:3
Step-by-Step Solution:
State variables are internal energy (U), Volume (V)
and Enthalpy (H).
Q 60ChemistrySection A • MCQp block elements
+4 / -1
Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because :
A
N2 is unreactive in the condition of atmosphere.
B
Oxides of nitrogen are unstable.
C
Reaction between them can occur in the presence of a catalyst.
D
The reaction is endothermic and require very high temperature.
Official Correct Answer:Option (D)
Step-by-Step Solution:
N2 is unreactive, its reaction with oxides is endothermic and require very high temperature.
N2+O2⇋(1483−2000K)2NO
(Endothermic and feasible at high temperature)
Q 61MathematicsSection A • MCQfunctions
+4 / -1
Let f(x)=ax2+bx+c be such that f(1)=3,f(−2)=λ and f(3)=4. If f(0)+f(1)+f(−2)+f(3)=14, then λ is equal to :
A
−4
B
213
C
223
D
4
Official Correct Answer:Option (D)
Step-by-Step Solution:
f(1)=a+b+c=3 ..... (i)
f(3)=9a+3b+c=4 .... (ii)
f(0)+f(1)+f(−2)+f(3)=14
OR c+3+(4a−2b+c)+4=14
OR 4a−2b+2c=7 ..... (iii)
From (i) and (ii) 8a+2b=1 ..... (iv)
From (iii) −(2)× (i)
⇒2a−4b=1 ..... (v)
From (iv) and (v) a=61,b=6−1 and c=3
f(−2)=4a−2b+c
=64+62+3=4
Q 62MathematicsSection A • MCQdifferential equations
+4 / -1
Let y=y(x) be the solution curve of the differential equation dxdy+x2−11y=(x+1x−1)1/2, x>1 passing through the point (2,31). Then 7y(8) is equal to :
Let the hyperbola H:a2x2−b2y2=1 pass through the point (22,−22). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?
A
(23,32)
B
(33,−62)
C
(3,−6)
D
(36,62)
Official Correct Answer:Option (B)
Step-by-Step Solution:
H:a2x2−b2y2=1
Focus of parabola : (ae,0)
Directrix : x=−ae.
Equation of parabola ≡y2=4aex
Length of latus rectum of parabola =4ae
Length of latus rectum of hyperbola =a2.b2
as given, 4ae=a2b2.e
2=a2b2 ...... (i)
∵ H passes through (22,−22)⇒a28−b28=1 ........ (ii)
From (i) and (ii) a2=4 and b2=8⇒e=3
⇒ Equation of parabola is y2=83x.
Q 64MathematicsSection B • Numericalprobability
+4 / -1
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100σ2 is equal to ________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 57
Official Correct Answer:57
Step-by-Step Solution:
X= Number of white ball drawn
P(X=0)=10C36C3=61
P(X=1)=10C36C2×4C1=21,
P(X=2)=10C36C1×4C2=103
and P(X=3)=10C36C0×4C3=301
Variance =σ2=∑PiXi2−(∑PiXi)2
σ2=21+1012+103−(21+106+101)2
=10056
100σ2=56.
Q 65MathematicsSection B • Numericalcomplex numbers
+4 / -1
Let z=a+ib,b=0 be complex numbers satisfying z2=zˉ⋅21−z. Then the least value of n∈N, such that zn=(z+1)n, is equal to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 6
Official Correct Answer:6
Step-by-Step Solution:
∵z2=z.21−∣z∣ ...... (1)
⇒∣z∣2=∣z∣.21−∣z∣
⇒∣z∣=21−∣z∣,
∵b=0⇒∣z∣=0
∴∣z∣=1 ...... (2)
∵z=a+ib then a2+b2=1 ...... (3)
Now again from equation (1), equation (2), equation (3) we get :
a2−b2+i2ab=(a−ib)20
∴a2−b2=a and 2ab=−b
∴a=−21 and b=±23
z=−21+23i or z=−21−23i
zn=(z+1)n⇒(zz+1)n=1
(1+z1)n=1
(21+3i)=1, then minimum value of n is 6.
Q 66MathematicsSection A • MCQapplication of derivatives
+4 / -1
The function f(x)=xex(1−x),x∈R, is :
A
increasing in (−21,1)
B
decreasing in (21,2)
C
increasing in (−1,−21)
D
decreasing in (−21,21)
Official Correct Answer:Option (A)
Step-by-Step Solution:
f(x)=xex(1−x),x∈R
f′(x)=xex(1−x).(1−2x)+ex(1−x)
=ex(1−x)[x−2x2+1]
=−ex(1−x)[2x2−x−1]
=−ex(1−x)(2x+1)(x−1)
∴f(x) is increasing in (−21,1) and decreasing in (−∞,−21)∪(1,∞)
Q 67MathematicsSection A • MCQdefinite integration
+4 / -1
Let In(x)=∫0x(t2+5)n1dt,n=1,2,3,…. Then :
A
50I6−9I5=xI5′
B
50I6−11I5=xI5′
C
50I6−9I5=I5′
D
50I6−11I5=I5′
Official Correct Answer:Option (A)
Step-by-Step Solution:
In(x)=0∫x(t2+5)n1dt
=0∫xI(t2+5)n1×IIIdt
=(t2+5)nt0x−0∫x(t2+5)n+1−2nt×tdt
=(x2+5)nx+0∫x2n((t2+5)n+1t2+5−5)dt
In(x)=(x2+5)nx+2nIn(x)−10nIn+1(x)
10nIn+1(x)−(2n−1)In(x)=xIn′(x)
For n=5
50I6(x)−9I5(x)=xI5′(x)
Q 68MathematicsSection A • MCQcircle
+4 / -1
Let the tangents at two points A and B on the circle x2+y2−4x+3=0 meet at origin O(0,0). Then the area of the triangle OAB is :
A
233
B
433
C
233
D
433
Official Correct Answer:Option (B)
Step-by-Step Solution:
x2+y2−4x+3=0
⇒(x−2)2+y2=1
AO=(OC)2−(AC)2
=4−1=3
sinθ=21⇒θ=6π
Also, AO=BO
Area of ΔOAB=21.OA.OBsin60∘
=21×3.3.23=433
Q 69MathematicsSection B • Numericalparabola
+4 / -1
Two tangent lines l1 and l2 are drawn from the point (2,0) to the parabola 2y2=−x. If the lines l1 and l2 are also tangent to the circle (x−5)2+y2=r, then 17r is equal to ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 9
Official Correct Answer:9
Step-by-Step Solution:
Given : y2=2−x
\eqalign{
& T \equiv y = mx - {1 \over {8m}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow (2,0) \cr}
⇒m2=161⇒m=±41
Tangents are y=41x−21,y=4−x+21
4y=x−2 and 4y+x=2
If these are also tangent to circle then dc=r
⇒175−2=r⇒r=(173)2
⇒17r=17.179=9
Q 70MathematicsSection A • MCQheight and distance
+4 / -1
A horizontal park is in the shape of a triangle OAB with AB=16. A vertical lamp post OP is erected at the point O such that ∠PAO=∠PBO=15∘ and ∠PCO=45∘, where C is the midpoint of AB. Then (OP)2 is equal to :
A
332(3−1)
B
332(2−3)
C
316(3−1)
D
316(2−3)
Official Correct Answer:Option (B)
Step-by-Step Solution:
OP=OAtan15=OBtan15 ...... (i)
OP=OCtan45⇒OP=OC ...... (ii)
OA=OB ....... (iii)
OC2+82=OA2
OP2+64=OP2(3−13+1)2
64=OP2[(3−1)2(3+1)2−(3−1)2]
=OP2((3−1)243)
OP2=4364(3−1)2=332(2−3)
Q 71MathematicsSection B • Numericalpermutations and combinations
+4 / -1
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then b+3g is equal to ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 17
Official Correct Answer:17
Step-by-Step Solution:
bC3.gC2=168
⇒6b(b−1)(b−2).2g(g−1)=168
⇒b(b−1)(b−2)g(g−1)=25.32.7
⇒b(b−1)(b−2)g(g−1)=6.7.8.3.2
∴b=8 and g=3
∴b+3g=17
Q 72MathematicsSection A • MCQ3d geometry
+4 / -1
A plane P is parallel to two lines whose direction ratios are −2,1,−3 and −1,2,−2 and it contains the point (2,2,−2). Let P intersect the co-ordinate axes at the points A,B,C making the intercepts α,β,γ. If V is the volume of the tetrahedron OABC, where O is the origin, and p=α+β+γ, then the ordered pair (V,p) is equal to :
A
(48,−13)
B
(24,−13)
C
(48,11)
D
(24,−5)
Official Correct Answer:Option (B)
Step-by-Step Solution:
Let a1=(−2,1,−3) and a2=(−1,2,−2)
Vector normal to plane n=a1×a2
n=(4,−1,−3)
Plane through (2,2,−2) and normal to n
(x−2,y−2,z+2).(4,−1,−3)=0
⇒4x−y−3z=12
⇒3x+−12y+−4z=1
Intercepts α, β, γ are 3,−12,−4
P=α+β+γ=−13
V=61×3×12×4=24
Q 73MathematicsSection A • MCQquadratic equation and inequalities
+4 / -1
Let α, β be the roots of the equation x2−2x+6=0 and α21+1,β21+1 be the roots of the equation x2+ax+b=0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)=0 are :
A
non-real complex numbers
B
real and both negative
C
real and both positive
D
real and exactly one of them is positive
Official Correct Answer:Option (B)
Step-by-Step Solution:
α+β=2, αβ=6
α21+1+β21+1=2+6α2+β2
=2+62−26=−a
(α21+1)(β21+1)=1+α21+β21+α2β21
=67+62−26=b
⇒a+b=6−5
So, equation is x2+617x+67=0
OR 6x2+17x+7=0
Both roots of equation are −ve and distinct
Q 74MathematicsSection A • MCQ3d geometry
+4 / -1
Let the lines
λx−1=1y−2=2z−3 and
−2x+26=3y+18=λz+28 be coplanar
and P be the plane containing these two lines.
Then which of the following points does NOT lie on P?
Checking the option gives (0, 4, 5) does not lie on the plane.
Q 75MathematicsSection B • Numericaldefinite integration
+4 / -1
The value of the integral 0∫2π60sinxsin(6x)dx is equal to _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 104
Official Correct Answer:104
Step-by-Step Solution:
I=0∫2π60.sinxsin6xdx
=60.20∫2π(3−4sin2x)(4cos2x−3)cosxdx
=1200∫2π(3−4sin2x)(1−4sin2x)cosxdx
Let sinx=t⇒cosxdx=dt
=1200∫1(3−4t2)(1−4t2)dt
=1200∫1(3−16t2+16t4)dt
=120[3t−316t3+516t5]01
=104
Q 76MathematicsSection A • MCQvector algebra
+4 / -1
Let S be the set of all a ∈R for which the angle between the vectors u=a(logeb)i^−6j^+3k^ and v=(logeb)i^+2j^+2a(logeb)k^, (b>1) is acute. Then S is equal to :
A
(−∞,−34)
B
Φ
C
(−34,0)
D
(712,∞)
Official Correct Answer:Option (B)
Step-by-Step Solution:
u=a(logeb)i−6j+3k
v=(logeb)i+2j+2a(logeb)k
For acute angle u.v>0
⇒a(logeb)2−12+6a(logeb)>0
∵b>1
Let logeb=t⇒t>0 as b>1
at2+6at−12>0∀t>0
⇒a∈ϕ
Q 77MathematicsSection B • Numericalellipse
+4 / -1
Let the tangents at the points P and Q on the ellipse 2x2+4y2=1 meet at the point R(2,22−2). If S is the focus of the ellipse on its negative major axis, then SP2+SQ2 is equal to ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 13
Let A and B be two events such that P(B∣A)=52,P(A∣B)=71 and P(A∩B)=91⋅ Consider
(S1) P(A′∪B)=65,
(S2) P(A′∩B′)=181
Then :
A
Both (S1) and (S2) are true
B
Both (S1) and (S2) are false
C
Only (S1) is true
D
Only (S2) is true
Official Correct Answer:Option (A)
Step-by-Step Solution:
P(A/B)=71⇒P(B)P(A∩B)=71
⇒P(B)=97
P(B/A)=52⇒P(A)P(A∩B)=52
P(A)=25.91=185
S2:P(A′∩B′)=181
S1: and P(A′∪B)=91+96+181=65.
Q 79MathematicsSection B • Numericalbinomial theorem
+4 / -1
Let the coefficients of the middle terms in the expansion of (61+βx)4,(1−3βx)2 and (1−2βx)6,β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then 50−β22d is equal to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 57
Official Correct Answer:57
Step-by-Step Solution:
Coefficients of middle terms of given expansions are 4C261β2,2C1(−3β),6C3(2−β)3 form an A.P.
∴2.2(−3β)=β2−25β3
⇒−24=2β−5β2
⇒5β2−2β−24=0
⇒5β2−12β+10β−24=0
⇒β(5β−12)+2(5β−12)=0
β=512
d=−6β−β2
∴50−β22d=50−2β2(−6β−β2)=50+β12+2=57
Q 80MathematicsSection A • MCQquadratic equation and inequalities
+4 / -1
Let S={x∈[−6,3]−{−2,2}:∣x∣−2∣x+3∣−1≥0} and
T={x∈Z:x2−7∣x∣+9≤0}.
Then the number of elements in S∩T is :
A
7
B
5
C
4
D
3
Official Correct Answer:Option (D)
Step-by-Step Solution:
∣x2∣−7∣x∣+9≤0
⇒∣x∣∈[27−13,27+13]
As x∈Z
So, x can be ±2,±3,±4,±5
Out of these values of x,
x=3,−4,−5
satisfy S as well
n(S∩T)=3
Q 81MathematicsSection B • Numericaltrigonometric functions and equations
+4 / -1
Let S=[−π,2π)−{−2π,−4π,−43π,4π}. Then the number of elements in the set ∣A={θ∈S:tanθ(1+5tan(2θ))=5−tan(2θ)} is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
Let tanα=5
∴tanθ=1+tanαtan2θtanα−tan2θ
∴tanθ=tan(α−2θ)
α−2θ=nπ+θ
⇒3θ=α−nπ
⇒θ=3α−3nπ;n∈Z
If θ∈[−π,π/2] then n=0,1,2,3,4 are acceptable
∴ 5 solutions.
Q 82MathematicsSection A • MCQlimits continuity and differentiability
+4 / -1
The function f:R→R defined by
f(x)=n→∞lim1+x2n+1−x2ncos(2πx)−x2nsin(x−1) is continuous for all x in :