Official NTA Question PaperFull Solved Paper2021Morning Shift
JEE Main 2021 17th March Morning 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 car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
A
(α+β)4αβt2
B
(α+β)2αβt2
C
2(α+β)αβt2
D
4(α+β)αβt2
Official Correct Answer:Option (C)
Step-by-Step Solution:
t1+t2=t,V′=0+αt1
V=u+at
0=αt1−βt2
t2=βαt1=t
t1=(α+ββ)t
Distance =21(t1+t2)×αt1 (area of triangle)
=21t×α(α+ββ)t
=2(α+β)αβt2
Q 2PhysicsSection A • MCQmagnetics
+4 / -1
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4π× 10−7 H/m)
A
πT
B
2 × 10−3π T
C
10−4π T
D
5π T
Official Correct Answer:Option (A)
Step-by-Step Solution:
B = μ n i
B = μrμ0 n i
B = 500 × 4π× 10−7× 103× 5
B = π× 10−3× 103
B = π T
Q 3PhysicsSection A • MCQray optics
+4 / -1
The thickness at the centre of a plane convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2 × 108 ms−1. The focal length of the lens is ____________.
A
0.30 cm
B
30 cm
C
15 cm
D
1.5 cm
Official Correct Answer:Option (B)
Step-by-Step Solution:
R2=32+(R−0.3)2
⇒R2=9+R2+0.09−2×0.3R
⇒2×0.3R=9.09
⇒R=15.15 cm
μ=VC = 2×1083×108 = 1.5
f1=(1.5−1)(R1)
f≃30 cm
Q 4PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A polyatomic ideal gas has 24 vibrational modes. What is the value of γ?
A
1.37
B
1.30
C
1.03
D
10.3
Official Correct Answer:Option (C)
Step-by-Step Solution:
f = 3T + 3R + 24V
= 30
γ = 1 + f2
γ = 1 + 302
= 1.066
Nearest Ans. = 1.03
Q 5PhysicsSection A • MCQheat and thermodynamics
+4 / -1
Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is :
A
K1+K22K1K2
B
K1K2K1+K2
C
2K1K2K1+K2
D
K1+K2K1K2
Official Correct Answer:Option (A)
Step-by-Step Solution:
Req=R1+R2
⇒Keq1A2l=K1Al+K2Al
⇒Keq2=K1l+K2l
⇒Keq2=K1K2K1+K2
⇒Keq=K1+K22K1K2
Q 6PhysicsSection A • MCQcircular motion
+4 / -1
A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μs, then the magnitude of negative lift FL acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity)
A
m(g−μsRv2)
B
−m(g+μsRv2)
C
m(μsRv2−g)
D
m(μsRv2+g)
Official Correct Answer:Option (C)
Step-by-Step Solution:
f=Rmv2
⇒μN=Rmv2
μ(mg+FL)=Rmv2
⇒mg+FL=Rμmv2
⇒FL=μRmv2−mg
=m(μRv2−g)
Q 7PhysicsSection A • MCQunits and measurements
+4 / -1
The vernier scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is ___________ cm. (least count = 0.01 cm)
A
8.58 cm
B
8.54 cm
C
8.56 cm
D
8.36 cm
Official Correct Answer:Option (B)
Step-by-Step Solution:
Reading = MSR + VSD × LC − zero error
Reading = 8.5 + 10(0.1)×6−100.2=8.54 cm
Q 8PhysicsSection B • Numericalsimple harmonic motion
+4 / -1
Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is Tb/Ta = x, where value of x is ___________. (Round off to the Nearest Integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
Ta=2πkM
Keqseries=k1+k2k1k2=2k
Tb=2πk/2M=2πk2M
Tb=2Ta
TaTb=2
∴ x = 2
Q 9PhysicsSection B • Numericalcommunication systems
+4 / -1
For VHF signal broadcasting, ___________ km2 of maximum service area will be covered by an antenna tower of height 30 m, if the receiving antenna is placed at ground. Let radius of the earth be 6400 km. (Round off to the Nearest Integer) (Take π as 3.14)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1206
Official Correct Answer:1206
Step-by-Step Solution:
d=2hR
area = πd2
Area = π(2hR) = 3.14 × 2 × 30 × 6400 × 103 . m2
= 1205.76 km2
≈ 1206 km2
Q 10PhysicsSection A • MCQatoms and nuclei
+4 / -1
If an electron is moving in the nth orbit of the hydrogen atom, then its velocity (vn) for the nth orbit is given as :
A
vn∝n1
B
vn∝ n2
C
vn∝ n
D
vn∝n21
Official Correct Answer:Option (A)
Step-by-Step Solution:
We know velocity of electron in nth shell of hydrogen atom is given by
v=nh2πkZe2
∴v∝n1
Q 11PhysicsSection A • MCQwork power and energy
+4 / -1
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
A
14.41 ms−1
B
19.0 ms−1
C
4.47 ms−1
D
1.00 ms−1
Official Correct Answer:Option (C)
Step-by-Step Solution:
K.E.f=5%KEi
21mv2=1005×21×m×202
v2=201×202=20
v=20=25 m/s
= 4.47 m/s
Q 12PhysicsSection B • Numericalrotational motion
+4 / -1
The following bodies,
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
a=(1+mR2I)gsinθ
IR = mR2, aR = g sinθ/2
ID = 2mR2, aD = 32 g sinθ
ISC = 2mR2, aSC = 32 g sinθ
ISC = 52mR2, aSS = 75 g sinθ
S = ut + 21at2,
t = a2S
∴ t ∝a1
solid sphere will take minimum time.
Q 13PhysicsSection B • Numericalelectromagnetic waves
+4 / -1
If 2.5 × 10−6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm2. (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 25
Official Correct Answer:25
Step-by-Step Solution:
Pressure = CIntensity (for absorbing surface)
I = P × C
I = 30cm22.5×10−6 N × 3 × 108 m/s
I = 25 W/cm2
Q 14PhysicsSection B • Numericalcapacitor
+4 / -1
A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelin plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _____________ pJ. (Assume no friction)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 864
Official Correct Answer:864
Step-by-Step Solution:
Ui=21cv2
=21×14×(12)2 pJ
= 1008 pJ
Uf=2kCQ2
=2×7×14(14×12)2
= 144 pJ
oscillating energy = Ui− Uf
= 1008 − 144
= 864 pJ
Q 15PhysicsSection A • MCQdual nature of radiation
+4 / -1
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
A
c1(E2m)1/2
B
c1(2mE)1/2
C
(2mE)1/2
D
c(2mE)1/2
Official Correct Answer:Option (B)
Step-by-Step Solution:
For photon, E = λhc
λp=Ehc .... (i)
For electron, λe=2mEhc .... (ii)
λpλe=Ehc2mEhc=2mc2E=c1(2mE)1/2
Q 16PhysicsSection A • MCQalternating current
+4 / -1
An AC current is given by I = I1 sinωt + I2 cosωt. A hot wire ammeter will give a reading :
When two soap bubbles of radii a and b (b > a) coalesce, the radius of curvature of common surface is :
A
abb−a
B
aba+b
C
a+bab
D
b−aab
Official Correct Answer:Option (D)
Step-by-Step Solution:
P1−P0=b4S ....(1)
P2−P0−a4S .....(2)
P2−P1=R4S ......(3)
eq(2) - eq(1) = eq(3)
⇒a1−b1=R1
∴R=b−aab
Q 18PhysicsSection B • Numericalcapacitor
+4 / -1
Four identical rectangular plates with length, l = 2 cm and breadth, b = 23 cm are arranged as shown in figure. The equivalent capacitance between A and C is dxε0. The value of x is ______________. (Round off to the Nearest Integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
Ceq=2C+C2C×C=32C
=32d∈0A=32×d∈0×(2×23)=d2ε0
Q 19PhysicsSection A • MCQcurrent electricity
+4 / -1
A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 × 10−3 ms−1. The number of free electrons in each cubic meter of the wire is ___________.
A
625 × 1025
B
1 × 1023
C
2 × 1025
D
2 × 106
Official Correct Answer:Option (A)
Step-by-Step Solution:
I=neAVd
n=eAVdI
=1.6×10−9×5×10−6×2×10−310
=161025=6.25×1027=625×1025
Q 20PhysicsSection A • MCQsimple harmonic motion
+4 / -1
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?
A
x = 2A
B
x = ± A
C
x = ±2A
D
x = 0
Official Correct Answer:Option (C)
Step-by-Step Solution:
KE = PE
21k(A2−X2)=21KX2
⇒A2−X2=X2
⇒2X2=A2
⇒X2=2A2
⇒X=±2A
Q 21PhysicsSection A • MCQheat and thermodynamics
+4 / -1
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
A
F1+F2n1F1T1+n2F2T2
B
n1F1+n2F2n1F1T1+n2F2T2
C
n1+n2n1T1+n2T2
D
n1+n2n1F1T1+n2F2T2
Official Correct Answer:Option (B)
Step-by-Step Solution:
Initial internal energy = Final internal energy
2F1n1RT1+2F2n2RT2=2F1n1RT+2F2n2RT
⇒T=F1n1+F2n2F1n1T1+F2n2T2
Q 22PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
A
2400 J
B
1600 J
C
1800 J
D
3200 J
Official Correct Answer:Option (A)
Step-by-Step Solution:
η=1−21
η=21
QW=η
Q1200=21
Q=2400J
Q 23PhysicsSection B • Numericalcurrent electricity
+4 / -1
The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
s=np
R1+R2=n[R1+R2R1R2]
⇒R12+R22+2R1R2=nR1R2
⇒R12+(2−n)R1R2+R22=0
For real roots, b2 - 4ac ≥ 0
[(2−n)R2]2−4×1×R22≥ 0
⇒(2−4)2R22≥4R22
⇒ 2 − n ≥±2
⇒ 2 − n ≥−2
⇒ n ≥ 4
So, minimum value for n = 4
Q 24PhysicsSection B • Numericalgravitation
+4 / -1
The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 64
Official Correct Answer:64
Step-by-Step Solution:
Ves=R2GM
VesR = const
∴Ves.R=10VesR′
⇒R′=100R = 64 km
Q 25PhysicsSection A • MCQrotational motion
+4 / -1
A triangular plate is shown. A force F = 4i− 3j is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :
A
− 15 + 203, 15 + 203
B
15 − 203, 15 + 203
C
15 + 203, 15 − 203
D
− 15 − 203, 15 − 203
Official Correct Answer:Option (D)
Step-by-Step Solution:
F=4i−3j
r1=5i+53j & r2=−5i+53j
Torque about 'O'
τO=r1×F=(−15−203)k=(15+203)(−k)
Torque about 'Q'
τQ=r2×F=(−15+203)k=(15−203)(−k)
Q 26PhysicsSection B • Numericalrotational motion
+4 / -1
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _____________. (Assuming the acceleration to be uniform).
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 728
Official Correct Answer:728
Step-by-Step Solution:
ωf=2460×602π
=82π
ωi=60900×2π=30π
α=tωf−ωi
=2682π−30π
= 2 π rad/sec2
θ=2αωf2−ωi2
=2×2π(82π+30π)(82π−30π)
=4π(112×52)π2
No. of revolution =2π(112×13)π
= 728
Q 27PhysicsSection A • MCQrotational motion
+4 / -1
A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity ω. The angular momentum of M about point A is LA which lies in the positive z direction and the angular momentum of M about point B is LB. The correct statement for this system is :
A
LA is constant, both in magnitude and direction
B
LB is constant in direction with varying magnitude
C
LB is constant, both in magnitude and direction
D
LA and LB are both constant in magnitude and direction
Official Correct Answer:Option (A)
Step-by-Step Solution:
Net force on M is towards A, hence torque is zero
about A.
LA=r×p
as r⊥p so LA = constant
Q 28PhysicsSection A • MCQatoms and nuclei
+4 / -1
Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?
A
8
B
6
C
1
D
4
Official Correct Answer:Option (B)
Step-by-Step Solution:
En=−13.6n2Z2
Enth of Carbon = E1st of Hydrogen
⇒−13.6×n262=−13.6×1212
⇒ n = 6
Q 29PhysicsSection B • Numericallaws of motion
+4 / -1
Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 73. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is ___________ N. (Round off to the Nearest Integer) [Take g as 9.8 ms−2]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 21
Official Correct Answer:21
Step-by-Step Solution:
a=M+mF
f = ma = mM+mF
mM+mF≤μmg (for no slipping)
⇒ F ≤μ(m + M)g
∴ Fmax = 73(0.5+4.5)×9.8 = 21 N
Q 30PhysicsSection A • MCQelectronic devices
+4 / -1
The output of the given combination gates represents :
A
AND Gate
B
NOR Gate
C
NAND Gate
D
XOR Gate
Official Correct Answer:Option (C)
Step-by-Step Solution:
Q 31ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
The mole fraction of a solute in a 100 molal aqueous solution is ___________ × 10−2. (Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 64
Official Correct Answer:64
Step-by-Step Solution:
Let weight of H2O = 1000 g
Moles of solute = 100
(mole)H2O = 181000
Mole fraction of solute = Totalmolesmoleofsolute
=100+181000100=28001800
Xsolute=64×10−2
Q 32ChemistrySection B • Numericalchemical equilibrium
+4 / -1
0.01 moles of a weak acid HA (Ka = 2.0 × 10−6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is __________ × 10−5 (Round off to the Nearest Integer).
[Neglect volume change on adding HA. Assume degree of dissociation <<1 ]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
Now,
Ka=[HA][H+][A−]
⇒2×10−6=10−2(0.1)(10−2α)
⇒α=2×10−5
Q 33ChemistrySection B • Numericalsolutions
+4 / -1
The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ × 10−5 mol dm−3. (Round off to the Nearest Integer).
[Given : Henry's law constant = KH = 8.0 × 104 kPa for O2. Density of water with dissolved oxygen = 1.0 kg dm−3 ]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 25
Official Correct Answer:25
Step-by-Step Solution:
Partial pressure (Pg) ∝ Solubility
Partial pressure (Pg) = KH× Solubility
20 × 103 = [8.0 × 104 × 103] × Solubility
Solubility = 8.0×10720×103 = 25 × 10-5
Q 34ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
The above reaction requires which of the following reaction conditions?
A
623 K, Cu, 300 atm
B
623 K, 300 atm
C
573 K, 300 atm
D
573 K, Cu, 300 atm
Official Correct Answer:Option (B)
Step-by-Step Solution:
Chlorobenzene is fused with NaOH at 623 K and
300 atmospheric pressure to get sodium
phenoxide.
Q 35ChemistrySection A • MCQcompounds containing nitrogen
+4 / -1
Hoffmann bromomide degradation of benzamide gives product A, which upon heating with CHCl3 and NaOH gives product B.
The structures of A and B are :
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 36ChemistrySection B • Numericalpractical organic chemistry
+4 / -1
In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ________%. (Round off to the Nearest Integer).
(Given atomic mass : C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 80
Official Correct Answer:80
Step-by-Step Solution:
Weight of (C6H5NO2)Theoretical = 783.9×123 = 6.15 g
% yield of reaction = WtheoreticalWactual×100=6.154.92×=80%
Q 37ChemistrySection A • MCQd and f block elements
+4 / -1
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?
A
zero
B
5.26
C
5.0
D
5.92
Official Correct Answer:Option (D)
Step-by-Step Solution:
25Mn = [Ar]3d5 4s2
Mn2+ = [Ar]3d5 4s0
No. of unpaired electron (n) = 5
μ=n(n+2)BM
=5×7=35=5.92 BM
Q 38ChemistrySection A • MCQs block elements
+4 / -1
The correct order of conductivity of ions in water is :
A
K+ > Na+ > Cs+ > Rb+
B
Cs+ > Rb+ > K+ > Na+
C
Na+ > K+ > Rb+ > Cs+
D
Rb+ > Na+ > K+ > Li+
Official Correct Answer:Option (B)
Step-by-Step Solution:
As the size of gaseous ion decreases, it get
more hydrated in water and hence, the size of
aqueous ion increases. When this bulky ion
move in solution, it experience greater
resistance and hence lower conductivity.
Q 39ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
The product ''A'' in the above reaction is :
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Ethylene glycol in presence of H+ will convert
ketone into cyclic ketal and the ester group remains
intact.
Q 40ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
Which of the following reaction is an example of ammonolysis?
A
C6H5COCl + C6H5NH2→ C6H5CONHC6H5
B
C6H5CH2CN ⟶\buildrel[H] C6H5CH2CH2NH2
C
C6H5CH2CN ⟶\buildrelHCl C6H5N+H3Cl−
D
C6H5CH2Cl + NH3→ C6H5CH2NH2
Official Correct Answer:Option (D)
Step-by-Step Solution:
Ammonolysis of alkyl halides is the reaction of an alkyl
halide with NH3 which leads to the preparation of
amines.
C6H5CH2Cl + NH3→ C6H5CH2NH2
Q 41ChemistrySection B • Numericalatomic structure
+4 / -1
A certain orbital has n = 4 and mL = −3. The number of radial nodes in this orbital is ____________. (Round off to the Nearest Integer).
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 0
Official Correct Answer:0
Step-by-Step Solution:
Number of radial nodes = n – ℓ – 1
n = 4, mL =–3 so ℓ = 3
radial nodes = 4 – 3 – 1 = 0
Q 42ChemistrySection A • MCQchemistry in everyday life
+4 / -1
With respect to drug-enzyme interaction, identify the wrong statement.
A
Allosteric inhibitor changes the enzyme's active site
B
Allosteric inhibitor competes with the enzyme's active site
C
Competitive inhibitor binds to the enzyme's active site
D
Non-Competitive inhibitor binds to the allosteric site
Official Correct Answer:Option (B)
Step-by-Step Solution:
Allosteric inhibitor combines at allosteric site and not the active
site.
Drugs inhibit the attachment of substrate on active site of enzymes
in two different ways :
(i) Drugs compete with the natural substrate for their attachment
on the active sites of enzymes. Such drugs are called
competitive inhibitors.
(ii) Some drugs do not bind to the enzyme's active site. These bind
to a different site of enzyme which is called allosteric site. This
binding of inhibitor at allosteric site changes the shape of the
active site in such a way that substrate cannot recognise it.This
is also called non-competitive inhibitor.
Non-competitive inhibitor changes the active site of enzyme
after binding at allosteric site.
Q 43ChemistrySection A • MCQchemical bonding and molecular structure
+4 / -1
A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is :
A
trigonal pyramidal
B
T-shaped
C
see-saw
D
planar triangular
Official Correct Answer:Option (B)
Step-by-Step Solution:
sp3d hybridised
T-shaped
Q 44ChemistrySection A • MCQperiodic classification of elements
+4 / -1
The absolute value of the electron gain enthalpy of halogens satisfies :
A
Cl > Br > F > I
B
Cl > F > Br > I
C
I > Br > Cl > F
D
F > Cl > Br > I
Official Correct Answer:Option (B)
Step-by-Step Solution:
The magnitude of electron gain enthalpy of halogen
atoms down the group show abnormal behavior.
The |ΔHeg| of F is lower than that of Cl due to its
smaller size. The incoming electron experiences
higher repulsive force due to valence electrons of
F than Cl. The correct order is Cl > F > Br > I.
Q 45ChemistrySection A • MCQenvironmental chemistry
+4 / -1
Reducing smog is a mixture of :
A
Smoke, fog and O3
B
Smoke, fog and N2O3
C
Smoke, fog and CH2 = CH − CHO
D
Smoke, fog and SO2
Official Correct Answer:Option (D)
Step-by-Step Solution:
Classical smog occurs in a cool humid climate, it is a mixture of smoke, fog, and SO2. It is known as reducing smog.
Q 46ChemistrySection A • MCQsurface chemistry
+4 / -1
A colloidal system consisting of a gas dispersed in a solid is called a/an :
A
foam
B
aerosol
C
solid sol
D
gel
Official Correct Answer:Option (C)
Step-by-Step Solution:
A colloidal system consisting of a gas dispersed in
a solid is called a ‘solid sol’.
Q 47ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2O72−. The molarity of the Fe2+ solution is __________ × 10-2 M. (Round off to the Nearest Integer).
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 24
Official Correct Answer:24
Step-by-Step Solution:
Cr2O72−+Fe2+⟶Cr3++Fe3+
Valance Factor of Cr2O72− = 6
Valance Factor of Fe2+ = 1
mili eq. of Cr2O72− = mili eq. of Fe2+
6(0.03 × 20) = 1(M × 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 × 10−2 M
Q 48ChemistrySection B • Numericalp block elements
+4 / -1
The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product 'A'. The reaction of 1 mol of 'A' with excess of AgNO3 in aqueous medium gives ___________ mol(s) of Ag. (Round off to the Nearest Integer).
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
Q 49ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
Mesityl oxide is a common name of :
A
3-Methyl cyclohexane carbaldehyde
B
4-Methyl pent-3-en-2-one
C
2,4-Dimethyl pentan-3-one
D
2-Methyl cyclohexanone
Official Correct Answer:Option (B)
Step-by-Step Solution:
Q 50ChemistrySection A • MCQd and f block elements
+4 / -1
Given below are two statements :
Statement I : Potassium permanganate on heating at 573K forms potassium manganate.
Statement II : Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature.
In the light of the above statements, choose the most appropriate answer from the options given below :
Q 51ChemistrySection B • Numericalchemical kinetics and nuclear chemistry
+4 / -1
For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ × 10−3 s−1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
k=t1ln[a−xa]
k=5702.303log(32100)
k=5702.303[log(102)−log25]
k=5702.303×0.5
k=2×10−3s−1
Q 52ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
Product ''A'' in the above chemical reaction is :
A
B
C
D
Official Correct Answer:Option (B)
Step-by-Step Solution:
Q 53ChemistrySection A • MCQionic equilibrium
+4 / -1
Which of the following compound CANNOT act as a Lewis base?
A
SF4
B
NF3
C
ClF3
D
PCl5
Official Correct Answer:Option (D)
Step-by-Step Solution:
NF3 has no vacant orbital neither in nitrogen nor in fluorine so it cannot
accept the electron & hence cannot acts as lewis acid and but for PCl5 P has
no L.P & hence it cannot acts as base but ClF3 (3 B.P + 2 L.P) & SF4 (4 B.P +
1 L.P)
Q 54ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
Given below are two statements :
Statement I : Retardation factor (Rf) can be measured in meter/centimeter.
Statement II : Rf value of a compound remains constant in all solvents.
Choose the most appropriate answer from the options given below :
A
Statement I is true but statement II is false
B
Statement I is false but statement II is true
C
Both statement I and statement II are true
D
Both statement I and statement II are false
Official Correct Answer:Option (D)
Step-by-Step Solution:
Retardation factor (Rf) is the ratio of the distance
moved by the substance from the baseline to the
distance moved by the solvent from the baseline.
So, it is dimensionless. The distance moved by the
substance is due to the adsorption of the substance on
the stationary phase. It does not depend on the
nature of the solvent. But the distance moved by the
solvent will change with the nature of the solvent.
Therefore, Rf will vary with the change in the solvent.
So, both the statements are false.
Q 55ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
Which of the following is an aromatic compound?
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
A compound which has (4n + 2)π electrons
completely delocalised over the cyclic ring is
aromatic.
Here 6 π electrons are present which follow the rule.
Q 56ChemistrySection B • Numericalthermodynamics
+4 / -1
The standard enthalpies of formation of Al2O3 and CaO are −1675 kJ mol-1 and −635 kJ mol−1 respectively.
For the reaction
3CaO + 2Al → 3Ca + Al2O3 the standard reaction enthalpy ΔrH0 = _________ kJ.
(Round off to the Nearest Integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 230
The point of intersection and sudden increase in the slope, in the diagram given below, respectively, indicates :
A
ΔG > 0 and decomposition of the metal oxide
B
ΔG < 0 and decomposition of the metal oxide
C
ΔG = 0 and reduction of the metal oxide
D
ΔG = 0 and melting or boiling point of the metal oxide
Official Correct Answer:Option (D)
Step-by-Step Solution:
From the Ellingham diagram given, the point of
intersection represents ΔG = 0 and the temperature
at which sudden increase in the slope occurs is
indicated by melting or boiling.
Q 58ChemistrySection A • MCQhydrogen
+4 / -1
The INCORRECT statement(s) about heavy water is (are)
(A) used as a moderator in nuclear reactor
(B) obtained as a by-product in fertilizer industry
(C) used for the study of reaction mechanism
(D) has a higher dielectric constant than water
Choose the correct answer from the options given below :
A
(D) only
B
(B) and (D) only
C
(B) only
D
(C) only
Official Correct Answer:Option (A)
Step-by-Step Solution:
Heavy water (D2O) is obtained as a by-product in
the fertilizer industry. It is used as a moderator in
nuclear reactors and for the study of the reaction
mechanisms. Its dielectric constant is lower than that
of H2O.
Q 59ChemistrySection A • MCQbiomolecules
+4 / -1
Which of the following is correct structure of tyrosine?
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Tyrosine is p-hydroxyphenylalanine. Its structure is
Q 60ChemistrySection B • Numericalgaseous state
+4 / -1
The pressure exerted by a non-reactive gaseous mixture of 6.4g of methane and 8.8 g of carbon dioxide in a 10L vessel at 27∘C is __________ kPa. (Round off to the Nearest Integer).
[Assume gases are ideal, R = 8.314 J mol−1 K−1
Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 150
Official Correct Answer:150
Step-by-Step Solution:
V = 10 L, T = 27∘ C = 300 K
(m)methane = 6.4 g, (m)CO2 = 8.8 g
PV = ntotalRT
P × 10 × 10−3 = (166.4+448.8)× 8.314 × 300
P × 10−2 = (0.4 + 0.2) × 8.314 × 300
P = 149652 Pa
P = 149.652 KPa ≈ 150 kPa
Q 61MathematicsSection A • MCQbinomial theorem
+4 / -1
If the fourth term in the expansion of (x+xlog2x)7 is 4480, then the value of x where x∈N is equal to :
A
3
B
1
C
4
D
2
Official Correct Answer:Option (D)
Step-by-Step Solution:
T4 = 7C3x4(xlog2x)3=4480
⇒35x4(xlog2x)3=4480
⇒x4(xlog2x)3=128
take log w.r.t. base 2 we get,
4log2x+3log2(xlog2x)=log2128
Let log2x=y
4y+3y2=7
⇒y=1,3−7
⇒log2x=1,3−7
⇒x=2,x=2−7/3
Q 62MathematicsSection A • MCQsets and relations
+4 / -1
In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?
A
Q and R
B
None of these
C
P and R
D
P and Q
Official Correct Answer:Option (B)
Step-by-Step Solution:
As none play all three games the intersection of all
three circles must be zero.
Hence none of P, Q, R justify the given statement
Q 63MathematicsSection B • Numerical3d geometry
+4 / -1
If the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3) is ax + by + cz − 7 = 0, then the value of 2a + b + c − 7 is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
Equation of plane can be written using family of planes : P1 + λP2 = 0
(2x − 7y + 4z − 3) + λ (3x − 5y + 4z + 11) = 0
It passes through (−2, 1, 3)
∴ (−4 + 7 + 12 − 3) + λ (−6 − 5 + 12 + 11) = 0
−2 + λ (12) = 0
λ = 61.
∴ 12x − 42y + 24z − 18 + 3x − 5y + 4z + 11 = 0
15x − 47y + 28z − 7 = 0
∴ a = 15, b = −47, c = 28
∴ 2a + b + c − 7 = 30 − 47 + 28 − 7 = 4
Q 64MathematicsSection A • MCQprobability
+4 / -1
Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :
The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is :
A
x + 3z = 0
B
3x − z = 0
C
x + 3z = 10
D
3x + z = 6
Official Correct Answer:Option (B)
Step-by-Step Solution:
Let the equation of the plane is a (x − 1) + b(y − 2) + c(z − 3) = 0
Y-axis lies on it.
D.R.'s of y-axis are 0, 1, 0
∴ 0.a + 1.b + 0.c = 0 ⇒ b = 0
∴ Equation of plane is a(x − 1) + c(z − 3) = 0
x = 0, z = 0 also satisfy it −a −3c = 0 ⇒ a = −3c
−3c (x − 1) + c (z − 3) = 0
−3 + 3 + z − 3 = 0
3x − z = 0
Q 66MathematicsSection A • MCQcircle
+4 / -1
The line 2x − y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x − 2y = 4. Then, the radius of the circle is :
A
53
B
45
C
35
D
54
Official Correct Answer:Option (C)
Step-by-Step Solution:
m1×m2=−1
a−22a−4−5×2=−1
a−2a−14=−1
a−14=2−a
2a=16
a = 8
∴ Centre (8, 2)
Radius = 36+9
=45
=35
Q 67MathematicsSection A • MCQfunctions
+4 / -1
The inverse of y=5logx is :
A
x=5logy
B
x=ylog51
C
x=5logy1
D
x=ylog5
Official Correct Answer:Option (B)
Step-by-Step Solution:
y=5logx
⇒logy=logx.log5
⇒logx=log5logy=log5y
⇒x=elog5y
⇒x=ylog5e
⇒x=ylog51
Q 68MathematicsSection B • Numericallimits continuity and differentiability
+4 / -1
If the function f(x)=x4cos(sinx)−cosx is continuous at each point in its domain and f(0)=k1, then k is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 6
Q 69MathematicsSection A • MCQstraight lines and pair of straight lines
+4 / -1
In a triangle PQR, the co-ordinates of the points P and Q are (−2, 4) and (4, −2) respectively. If the equation of the perpendicular bisector of PR is 2x − y + 2 = 0, then the centre of the circumcircle of the ΔPQR is :
A
(−1, 0)
B
(1, 4)
C
(0, 2)
D
(−2, −2)
Official Correct Answer:Option (D)
Step-by-Step Solution:
Mid point of PQ≡(2−2+4,24−2)≡(1,1)
Slope of PQ=−2−44+2=−1
Slope of perpendicular bisector of PQ = 1
Equation of perpendicular bisector of PQ
y−1=1(x−1)
⇒y=x
Solving with perpendicular bisector of PR,
2x − y + 2 = 0
Circumcentre is (−2, −2)
Q 70MathematicsSection A • MCQdefinite integration
+4 / -1
Which of the following statements is correct for the function g(α) for α∈ R such that
Q 73MathematicsSection A • MCQdifferential equations
+4 / -1
Which of the following is true for y(x) that satisfies the differential equation
dxdy = xy − 1 + x − y; y(0) = 0 :
A
y(1) = 1
B
y(1) = e−21− 1
C
y(1) = e21− e−21
D
y(1) = e21− 1
Official Correct Answer:Option (B)
Step-by-Step Solution:
dxdy=(x−1)y+(x−1)
dxdy=(x−1)(y+1)
y+1dy=(x−1)dx
Integrating both sides, we get
ln(y+1)=2x2−x+c
x=0,y=0
⇒c=0
∴ln(y+1)=2x2−x
putting x=1,ln(y+1)=21−1=−21
y+1=e−21
y=e−21−1
∴y(1)=e−21−1
Q 74MathematicsSection B • Numericalstraight lines and pair of straight lines
+4 / -1
The maximum value of z in the following equation z = 6xy + y2, where 3x + 4y ≤ 100 and 4x + 3y ≤ 75 for x ≥ 0 and y ≥ 0 is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 904
Official Correct Answer:904
Step-by-Step Solution:
3x + 4y ≤ 100
4x + 3y ≤ 75
x ≥ 0, y ≥ 0
Feasible region is shown in the graph
Let maximum value of 6xy + y2 = c
For a solution with feasible region,
6xy + y2 = c and 4x + 3y = 75 must have at least one positive solution.
y2+6y(475−3y)−c=0
⇒27y2−2225y+c=0
⇒(2225)2≥4.27.c
⇒c≤562252≈904
Q 75MathematicsSection A • MCQcircle
+4 / -1
Choose the incorrect statement about the two circles whose equations are given below :
x2 + y2− 10x − 10y + 41 = 0 and
x2 + y2− 16x − 10y + 80 = 0
A
Distance between two centres is the average of radii of both the circles.
B
Both circles pass through the centre of each other.
C
Circles have two intersection points.
D
Both circle's centers lie inside region of one another.
Official Correct Answer:Option (D)
Step-by-Step Solution:
S1≡ x2 + y2− 10x − 10y + 41 = 0
Centre C1≡ (5, 5), radius r1 = 3
S2≡ x2 + y2− 16x − 10y + 80 = 0
Centre C2≡ (8, 5), radius r2 = 3
Distance between centres = 3
Hence both circles pass through the centre of each other, have two intersection point and distance between two centres in average of radii of both the circles.
Hence, option (d) is the incorrect statement.
Q 76MathematicsSection A • MCQcomplex numbers
+4 / -1
The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :
A
1
B
21| z |2
C
21| z + iz |2
D
21
Official Correct Answer:Option (B)
Step-by-Step Solution:
Each side length = |z|
Area of Δ = 21 (area of square)
= 21|z|2
Q 77MathematicsSection B • Numericalcircle
+4 / -1
The minimum distance between any two points P1 and P2 while considering point P1 on one circle and point P2 on the other circle for the given circles' equations
x2 + y2− 10x − 10y + 41 = 0
x2 + y2− 24x − 10y + 160 = 0 is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
S1:(x−5)2+(y−5)2=9
Centre (5, 5), r1 = 3
S2:(x−12)2+(y−5)2=9
Centre (12, 5), r2 = 3
So (P1P2)min = 1
Q 78MathematicsSection B • Numericalprobability
+4 / -1
Let there be three independent events E1, E2 and E3. The probability that only E1 occurs is α, only E2 occurs is β and only E3 occurs is γ. Let 'p' denote the probability of none of events occurs that satisfies the equations (α− 2β)p = αβ and (β− 3γ)p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1).
Then, ProbabilityofoccurrenceofE3ProbabilityofoccurrenceofE1 is equal to _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 6
Q 81MathematicsSection B • Numericaldifferentiation
+4 / -1
If f(x)=sin(cos−1(1+22x1−22x)) and its first derivative with respect to x is −abloge2 when x = 1, where a and b are integers, then the minimum value of | a2− b2 | is ____________ .
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 481
Official Correct Answer:481
Step-by-Step Solution:
f(x)=sincos−1(1+(2x)21−(2x)2)
=sin(2tan−12x)
f′(x)=cos(2tan−12x).2.1+(2x)21×2x.loge2
∴f′(1)=cos(2tan−12).1+42×2×loge2
⇒f′(1)=coscos−1(1+221−22).54loge2
=−2512loge2
⇒a=25,b=12
∣a2−b2∣=∣625−144∣=481
Q 82MathematicsSection A • MCQpermutations and combinations
+4 / -1
Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to :
A
5
B
2
C
4
D
6
Official Correct Answer:Option (C)
Step-by-Step Solution:
Total matches between boys of both team = 7C1×4C1=28
Total matches between girls of both team = nC16C1=6n
Now, 28 + 6n = 52
⇒ n = 4
Q 83MathematicsSection A • MCQmathematical reasoning
+4 / -1
If the Boolean expression (p ⇒ q) ⇔ (q * (∼p) is a tautology, then the boolean expression (p * (∼q)) is equivalent to :
A
q ⇒ p
B
p ⇒ q
C
p ⇒∼ q
D
∼q ⇒ p
Official Correct Answer:Option (A)
Step-by-Step Solution:
The Boolean expression
(p⇒q)⇔(q∗(∼p)) is a tautology.
Making the truth table for this
p
q
p→q
q∗∼p
∼q
−q∧p
∼(∼q∧p)
T
T
T
T
F
F
T
T
F
F
F
T
T
F
F
T
T
T
F
F
T
F
F
T
T
T
F
T
∴∼(∼q∧p)=q∨∼p=∼p∨q
∴∗ is equivalent to v.
So, p∗∼q=p∨∼q=∼q∨p
=q⇒p
Q 84MathematicsSection B • Numericalbinomial theorem
+4 / -1
If (2021)3762 is divided by 17, then the remainder is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4