Official NTA Question PaperFull Solved Paper2023Morning Shift
JEE Main 2023 25th January 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.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for ∣Vz∣>±v≥∣v0∣ where v0 is the threshold voltage and Vz is the breakdown voltage.
In the light of the above statements, choose the correct answer from the options given below
A
Both A and R are true but R is NOT the correct explanation of A
B
A is true but R is false
C
Both A and R are true and R is the correct explanation of A
D
A is false but R is true
Official Correct Answer:Option (D)
Step-by-Step Solution:
Photodiodes are used in reverse bias therefore the assertion is incorrect.
Q 2PhysicsSection A • MCQheat and thermodynamics
+4 / -1
The root mean square velocity of molecules of gas is
A
Proportional to temperature (T)
B
Inversely proportional to square root of temperature (T1)
C
Proportional to square of temperature (T2)
D
Proportional to square root of temperature (T)
Official Correct Answer:Option (D)
Step-by-Step Solution:
The rms speed of a gas molecule is
VRMS=M3RT
VRMS∝T
Q 3PhysicsSection A • MCQcapacitor
+4 / -1
A parallel plate capacitor has plate area 40 cm2 and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :
A
10ε0F
B
24ε0F
C
103ε0F
D
310ε0F
Official Correct Answer:Option (D)
Step-by-Step Solution:
This can be seen as two capacitors in series combination so
Choose the correct answer from the options given below :
A
A-III, B-IV, C-I, D-II
B
A-II, B-I, C-IV, D-III
C
A-III, B-I, C-IV, D-II
D
A-I, B-III, C-IV, D-II
Official Correct Answer:Option (C)
Step-by-Step Solution:
A→B0=4πr−μ0I+2rμ0I−4πrμ0I
B0=2πrμ0I(π−1)A→ III
B→B0=4πrμ0I+4rμ0I+4πrμ0I
B0=4πrμ0I(π+2)B→I
C→B0=4πrμ0I+4rμ0I+0
B0=4πrμ0I(π+1)C→IV
D. →B0=4rμ0ID→II
Q 6PhysicsSection B • Numericalwork power and energy
+4 / -1
An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ=kx, where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E=knsinθ. The value of n is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 × 10−3 kg, density is 8.92 × 103 kg/m3 and resistivity is 1.7 × 10−8Ω-m. The length of wire is :
A
l=100 m
B
l=6.8 m
C
l=5 m
D
l=10 m
Official Correct Answer:Option (D)
Step-by-Step Solution:
m=8.92×10−3kg
Density =8.92×103kg/m3
Volume =8.92×1038.92×10−3=(10−6)m3
Resistance =23.4=1.7Ω=(Aρl)
1.7=(Al)ρl2
⇒1.7=10−61.7×10−8×l2
l2=100
l=10m
Q 8PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :
A
300 K
B
1000 K
C
900 K
D
360 K
Official Correct Answer:Option (B)
Step-by-Step Solution:
η=1−Tsource Tsink
50% efficiency ⇒21=1−Tsource Tsink
21=1−600Tsink ⇒Tsink =300
Now, 70% efficiency ⇒107=1−Tsource Tsink
Tsource 300=103
Tsource =1000K
Q 9PhysicsSection B • Numericalatoms and nuclei
+4 / -1
The wavelength of the radiation emitted is λ0 when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will x20λ0. The value of x is _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 27
Official Correct Answer:27
Step-by-Step Solution:
Second excited state → first excited state , n=3 to n=2
λ01=R(41−91)=(365R)
Third excited state → second orbit , n=4 to n=2
λ1=R(41−161)=(163R)
Taking ratio of (1) and (2)
λ0λ=365×316=(2720)
λ=2720λ0
x=27
Q 10PhysicsSection A • MCQcommunication systems
+4 / -1
A message signal of frequency 5 kHz is used to modulate a carrier signal of frequency 2 MHz. The bandwidth for amplitude modulation is :
A
10 kHz
B
2.5 kHz
C
20 kHz
D
5 kHz
Official Correct Answer:Option (A)
Step-by-Step Solution:
Given
Signal frequency fm = 5kHz
Carrier wave frequency fc=2MHzfc=2000KHz
The resultant signal will have band width of frequency given by
The ratio of the density of oxygen nucleus (816O) and helium nucleus (24He) is
A
4 : 1
B
1 : 1
C
2 : 1
D
8 : 1
Official Correct Answer:Option (B)
Step-by-Step Solution:
Nuclear density is independent of mass number
As nuclear density =34πR3Au
Also, R=R0A31
And R3=R03A
⇒ Nuclear density =34πR03AAu
Nuclear density =4πR033u
⇒ Nuclear density is independent of A
Q 12PhysicsSection A • MCQdual nature of radiation
+4 / -1
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ0. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :
A
3 λ0
B
9 λ0
C
2λ0
D
2λ0
Official Correct Answer:Option (C)
Step-by-Step Solution:
When electron is accelerated through potential difference V, then
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :
A
60 μm
B
48 μm
C
36 μm
D
12 μm
Official Correct Answer:Option (A)
Step-by-Step Solution:
In Young's double-slit experiment, the distance between the slits and the screen, L=1 m, the wavelength of the light, λ=600 nm=600×10−9 m, and the position of the 5th bright fringe from the central maximum, y=5 cm=0.05 m.
The distance between the central maximum and the nth bright fringe is given by the formula:
yn=dnλL
where d is the distance between the two slits.
We can rearrange this formula to solve for d:
d=ynnλL
Substituting the values given in the question, we get:
d=55×600×10−9×1×100
d=6×10−5 m
d=60μm
Therefore, the separation between the slits is 60μm.
Q 14PhysicsSection A • MCQcircular motion
+4 / -1
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g = 10 m/s2)
A
2π
B
6π
C
4π
D
3π
Official Correct Answer:Option (C)
Step-by-Step Solution:
In car’s frame, FBD of bob
where aP= Pseudoforce or centrifugal force
θ=tan−1(gaP)=tan−1(Rgv2)=tan−1(40×10400)=45∘
Q 15PhysicsSection B • Numericalvector algebra
+4 / -1
If P=3i+3j+2k and Q=4i+3j+2.5k then, the unit vector in the direction of P×Q is x1(3i+j−23k). The value of x is _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
P=3i^+3j^+2k^
Q=4i^+3j^+2.5k^
P×Q=i^34j^33k^22.5
=i^(23)−j^(−21)+k^(−3)
=23i^+2j^−3k^
∣P×Q∣=43+41+3=2
Unit vector along P×Q=41(3i^+j^−23k^)
x=4
Q 16PhysicsSection B • Numericalproperties of matter
+4 / -1
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45∘ with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x×104 Nm−2. 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:
Given, Length, L=62.8cm
diameter, d=4mm
radius, r=2mm
y=x×104N/m2
According to graph, Δl/f=tan45∘=1
Y=A⋅ΔLFL
Y=3.14×2×2×10−61×62.8×10−2=5×104N/m2
∴ x = 5
Q 17PhysicsSection B • Numericalwaves
+4 / -1
The distance between two consecutive points with phase difference of 60∘ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is travelling is __________ km/s
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 18
A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is ___________ (in degree).
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 45
Official Correct Answer:45
Step-by-Step Solution:
0.5e=21mvx2⇒vx=me Along xL=vxt=met Along y vy=meEtdividingLvy=Eme=Evx⇒tanθ=vxvy=E×L=10×0.1=1θ=45∘
Q 19PhysicsSection A • MCQmotion
+4 / -1
A car travels a distance of 'x' with speed v1 and then same distance 'x' with speed v2 in the same direction. The average speed of the car is :
A
2(v1+v2)v1v2
B
v1+v22v1v2
C
v1+v22x
D
2v1+v2
Official Correct Answer:Option (B)
Step-by-Step Solution:
Average velocity = Total time Total displacement =v1x+v2xx+x=v1+v22v1v2
Q 20PhysicsSection A • MCQalternating current
+4 / -1
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency ω0. The value of x is :
A
1/4
B
1/16
C
4
D
16
Official Correct Answer:Option (A)
Step-by-Step Solution:
The resonance frequency of LC oscillations circuit is
ω0=LC1L→2LC→8Cω=2L×8C1=4LC1ω=4ω0
So x=41
Q 21PhysicsSection A • MCQmagnetic properties of matter
+4 / -1
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is :
A
1Am−1
B
2.4×10−3Am−1
C
1.2×103Am−1
D
2.4×103Am−1
Official Correct Answer:Option (C)
Step-by-Step Solution:
Number of turns per unit length =21200=600
So, Magnetic Intensity H=nI
=600×2Am−1=1200Am−1
Q 22PhysicsSection A • MCQproperties of matter
+4 / -1
A bowl filled with very hot soup cools from 98∘C to 86∘C in 2 minutes when the room temperature is 22∘C. How long it will take to cool from 75∘C to 69∘C?
A
2 minutes
B
0.5 minute
C
1.4 minutes
D
1 minute
Official Correct Answer:Option (C)
Step-by-Step Solution:
From Newton's law of cooling.
dtdT=−k(T−Ts)
Case I:dT=12∘C,dt=2min
212=−k[92−22∘]=−k70
Case II : dT=6∘C
dt6=−k[72−22]=−k50
From (1) and (2)
dt=1.4min
Q 23PhysicsSection A • MCQrotational motion
+4 / -1
An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s2)
A
90 N
B
240 N
C
30 N
D
300 N
Official Correct Answer:Option (D)
Step-by-Step Solution:
Torque balance about ' O '
2T×0.6=20×0.5+80×1
T×0.3=10+80=90
T=3900=300N
Q 24PhysicsSection A • MCQelectromagnetic waves
+4 / -1
An electromagnetic wave is transporting energy in the negative z direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be the direction of the magnetic field of the wave at that point and instant?
A
Negative direction of y
B
Positive direction of z
C
Positive direction of x
D
Negative direction x
Official Correct Answer:Option (C)
Step-by-Step Solution:
As, poynting vector
S=E×H
Given energy transport = negative z direction
Electric field = positive y direction
(−k^)=(+j^)×[i^]
Hence according to vector cross product magnetic field should be positive x direction.
Q 25PhysicsSection B • Numericalalternating current
+4 / -1
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.
(Take π2=10)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 100
Official Correct Answer:100
Step-by-Step Solution:
∵ For maximum amplitude of current, circuit should be at resonance.
T is the time period of simple pendulum on the earth's surface. Its time period becomes x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be :
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)
(Take g = 10 m s−2 , radius of earth = 6400 km)
A
12 hours
B
1 hour 24 minutes
C
24 hours
D
1 hour 40 minutes
Official Correct Answer:Option (B)
Step-by-Step Solution:
Gravitational acceleration at a distance of r from centre of earth is given by
g′=Rgr
Where R is the radius of earth
So, dt2d2r=−Rgr
⇒T=2πgR=2π106400000
=2π×800sec
=5024sec
= 1 hour 24 minutes (approx.)
Q 28PhysicsSection B • Numericalrotational motion
+4 / -1
ICM is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IAB is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 32R from center. Where R is the radius of the disc. The ratio of IAB and ICM is x:9. The value of x is _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 17
Official Correct Answer:17
Step-by-Step Solution:
IAB=Icm+M×(32R)2
=21MR2+94MR2
=18(9+8)MR2=(1817)MR2
IcmIAB=1/217/18=(917)
Value of x=17
Q 29PhysicsSection B • Numericalcurrent electricity
+4 / -1
In the given circuit, the equivalent resistance between the terminal A and B is __________ Ω.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 10
Official Correct Answer:10
Step-by-Step Solution:
Both 4Ω resistance and left 2Ω resistance get short-circuited. So, no current will flow through them.
Remove the resistors that have no current. Now the equivalent circuit becomes -
Req=3+(2∥2)+6Req=3+1+6Req=10Ω
Q 30PhysicsSection B • Numericalray optics
+4 / -1
A ray of light is incident from air on a glass plate having thickness 3 cm and refractive index 2. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____________ × 10−2 cm. (given sin15∘=0.26)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 52
Official Correct Answer:52
Step-by-Step Solution:
sini=21=45∘
⇒ at point (1)
μsinr=sini=21
sinr=21⇒r=30∘
Lateral displacement
=cosrtsin(15∘)=(23)3×0.26
=2×0.26
=0.52cm=52×10−2cm
Q 31ChemistrySection A • MCQbiomolecules
+4 / -1
Match items of Row I with those of Row II.
Row I :
Row II :
(i) α-D-(−)-Fructofuranose
(ii) β-D-(−)-Fructofuranose
(iii) α-D-(−) Glucopyranose
(iv) β-D-(−)-Glucopyranose
Correct match is
A
A → iii, B → iv, C → i, D → ii
B
A → iv, B → iii, C → i, D → ii
C
A → i, B → ii, C → iii, D → iv
D
A → iii, B → iv, C → ii, D → i
Official Correct Answer:Option (A)
Step-by-Step Solution:
(i) α-D-(-)-Fructofuranose -
(ii) β-D-(-)-Fructofuranose -
(iii) α-D-(-)-Glucopyranose -
(iv) β-D-(-)-Glucopyranose-
Q 32ChemistrySection A • MCQp block elements
+4 / -1
Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively :
An athlete is given 100 g of glucose (C6H12O6) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ____________ g (Nearest integer)
Assume that there is no other way of consuming stored energy.
Given : The enthalpy of evaporation of water is 45 kJ mol−1
Molar mass of C, H & O are 12, 1 and 16 g mol−1.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 360
Official Correct Answer:360
Step-by-Step Solution:
C6H12O6(s)+6O2→6CO2(g)+6H2O(l)
Extra energy used to convert H2O(l) into H2O(l) into H2O(g)
Which one of the following reactions does not occur during extraction of copper?
A
2FeS+3O2→2FeO+2SO2
B
FeO+SiO2→FeSiO3
C
CaO+SiO2→CaSiO3
D
2Cu2s+3O2→2Cu2O+2SO2
Official Correct Answer:Option (C)
Step-by-Step Solution:
CuFeS2+O2⟶ Partial roasting Cu2S+FeO+SO2+ very small FeS+ very small Cu2OCu2S+O2→Cu2O+SO2FeS+O2→FeO+SO2FeO+SiO2→FeSiO3
No formation of calcium silicate (CaSiO3) in extraction of Cu.
Q 35ChemistrySection A • MCQpractical organic chemistry
+4 / -1
Match the List-I with List-II :
List-I Cations
List-II Group reagents
A.
Pb2+,Cu2+
i)
H2S gas in presence of dilute HCl
B.
Al3+,Fe3+
ii)
(NH4)2CO3 in presence of NH4OH
C.
Co2+,Ni2+
iii)
NH4OH in presence of NH4Cl
D.
Ba2+,Ca2+
iv)
H2S in presence of NH4OH
Correct match is -
A
A → i, B → iii, C → iv, D → ii
B
A → i, B → iii, C → ii, D → iv
C
A → iii, B → i, C → iv, D → ii
D
A → iv, B → ii, C → iii, D → i
Official Correct Answer:Option (A)
Step-by-Step Solution:
Cations
Group No.
Group reagent
Pb+2,Cu+2
II
H2S(g) in presence of dil HCl
Al+3,Fe+3
III
nbsp;nbsp;nbsp;nbsp;NH4Clnbsp;nbsp;NH4OH in presence of
CO+2,Ni+2
IV
nbsp;nbsp;nbsp;nbsp;NH4OHnbsp;nbsp;H2S in presence of
Ba+2,Ca+2
V
nbsp;nbsp;nbsp;nbsp; of NH4OHnbsp;nbsp;(NH4)2CO3 in presence
Q 36ChemistrySection A • MCQperiodic classification of elements
+4 / -1
Inert gases have positive electron gain enthalpy. Its correct order is :
A
He < Ne < Kr < Xe
B
He < Xe < Kr < Ne
C
Xe < Kr < Ne < He
D
He < Kr < Xe < Ne
Official Correct Answer:Option (B)
Step-by-Step Solution:
Electron textgain
He
Ne
Ar
Kr
Xe
Enthalpyl /mathrmkJmol−1
48
116
96
96
77
Q 37ChemistrySection A • MCQsolid state
+4 / -1
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at 31rd of the total faces. The empirical formula of the compound is :
A
X2Y1.5
B
X3Y2
C
XY2.5
D
X2.5Y
Official Correct Answer:Option (B)
Step-by-Step Solution:
Number of X particles =4×81+1=1.5 Number of Y particles =6×31×21=1∴ Empirical formula =X1.5Y1=X3Y2
Q 38ChemistrySection A • MCQpractical organic chemistry
+4 / -1
Match List I with List II
List I Elements
List II Colour imparted to the flame
A.
K
I.
Brick Red
B.
Ca
II.
Violet
C.
Sr
III.
Apple Green
D.
Ba
IV.
Crimson Red
Choose the correct answer from the options given below :
A
A-II, B-IV, C-I, D-III
B
A-II, B-I, C-IV, D-III
C
A-II, B-I, C-III, D-IV
D
A-IV, B-III, C-II, D-I
Official Correct Answer:Option (B)
Step-by-Step Solution:
Element
nbsp;nbsp;nbsp;nbsp; the flame nbsp;nbsp; Colour imparted to
(A)
K
Violet
(B)
Ca
Brick red
(C)
Sr
Crimson red
(D)
Ba
Apple green
Q 39ChemistrySection A • MCQalcohols phenols and ethers
+4 / -1
In the cumene to phenol preparation in presence of air, the intermediate is
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Q 40ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ____________ (Nearest Integer)
(Given : Atomic mass Ba: 137 u, S: 32 u, O: 16 u)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 42
Official Correct Answer:42
Step-by-Step Solution:
% of sulphur =23332× weight of organic compound weight of BaSO4 formed ×100=23332×0.4711.4439×100=42.10
Q 41ChemistrySection A • MCQp block elements
+4 / -1
Compound A reacts with NH4Cl and forms a compound B. Compound B reacts with H2O and excess of CO2 to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively :
Q 43ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
Identify the product formed (A and E)
A
B
C
D
Official Correct Answer:Option (C)
Step-by-Step Solution:
Q 44ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
The correct order in aqueous medium of basic strength in case of methyl substituted amines is :
A
NH3>Me3N>MeNH2>Me2NH
B
Me3N>Me2NH>MeNH2>NH3
C
Me2NH>MeNH2>Me3N>NH3
D
Me2NH>Me3N>MeNH2>NH3
Official Correct Answer:Option (C)
Step-by-Step Solution:
In aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting H+. After considering all these factors overall basic strength order is
Me2NH>MeNH2>Me3N>NH3
Q 45ChemistrySection A • MCQsome basic concepts of chemistry
+4 / -1
'25 volume' hydrogen peroxide means
A
1 L marketed solution contains 75 g of H2O2.
B
1 L marketed solution contains 250 g of H2O2.
C
1 L marketed solution contains 25 g of H2O2.
D
100 mL marketed solution contains 25 g of H2O2.
Official Correct Answer:Option (A)
Step-by-Step Solution:
Molarity of H2O2 soln =11.2 volume strength
=11.225=2.23M
∴ amount of H2O2 in one litre =2.23×34=75gm
Q 46ChemistrySection A • MCQsurface chemistry
+4 / -1
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph :
A
(a)
B
(b)
C
(d)
D
(c)
Official Correct Answer:Option (D)
Step-by-Step Solution:
The correct plot for enzyme catalysed reaction is
Q 47ChemistrySection B • Numericalchemical bonding and molecular structure
+4 / -1
The total number of lone pairs of electrons on oxygen atoms of ozone is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 6
Official Correct Answer:6
Step-by-Step Solution:
Total no, of lone pairs on oxygen atoms = 6
Q 48ChemistrySection A • MCQhydrocarbons
+4 / -1
The correct sequence of reagents for the preparation of Q and R is :
Q 54ChemistrySection A • MCQenvironmental chemistry
+4 / -1
Some reactions of NO2 relevant to photochemical smog formation are
Identify A, B, X and Y :
A
X=[O],Y=NO,A=O2,B=O3
B
X=N2O,Y=[O],A=O3,B=NO
C
X=21O2,Y=NO2,A=O3,B=O2
D
X=NO,Y=[O],A=O2,B=N2O3
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 55ChemistrySection B • Numericalionic equilibrium
+4 / -1
A litre of buffer solution contains 0.1 mole of each of NH3 and NH4Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ × 10−3 (Nearest integer)
[Given : pKb(NH3)=4.745
log2=0.301
log3=0.477
T=298K]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 9079
The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ × 10−2 mL (Nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 12
Official Correct Answer:12
Step-by-Step Solution:
millimole of NaOH=0.24×25∴ millimole of acid =0.24×25⇒ mass of acid =0.24×25×24.2mg for pure acid, V=dw;(d=1.21kg/L=1.21g/ml)∴V=1.120.24×25×24.2×10−3=120×10−3ml=12×10−2ml
Q 57ChemistrySection A • MCQchemistry in everyday life
+4 / -1
Which of the following statements is incorrect for antibiotics?
A
An antibiotic should promote the growth or survival of microorganisms.
B
An antibiotic must be a product of metabolism.
C
An antibiotic should be effective in low concentrations.
D
An antibiotic is a synthetic substance produced as a structural analogue of naturally occurring antibiotic.
Official Correct Answer:Option (A)
Step-by-Step Solution:
Statement (A) is incorrect. Antibiotics are substances that are produced by microorganisms to inhibit the growth of other microorganisms. Antibiotics are not meant to promote the growth or survival of microorganisms, but rather to control or eliminate them.
Q 58ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH− is
A
B
C
D
Official Correct Answer:Option (B)
Step-by-Step Solution:
Aryl halides having E.W.G at O-or P-position have greater rate than the m-isomers towards nucleophilic aromatic substitution.
Hence the correct answer is (B)
Q 59ChemistrySection A • MCQalcohols phenols and ethers
+4 / -1
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Acetal / Ketal is stable in basic medium.
Reason R : The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium.
In the light of the above statements, choose the correct answer from the options given below :
A
A is false but R is true.
B
Both A and R are true but R is NOT the correct explanation of A
C
Both A and R are true and R is the correct explanation of A
D
A is true but R is false.
Official Correct Answer:Option (D)
Step-by-Step Solution:
For Assertion: Acetal and ketals are basically
ethers hence they must be stable in basic medium
but should break down in acidic medium.
Hence assertion is correct.
For reason: Alkoxide ion (RO–) is not considered a good leaving group hence reason must be false.
Q 60ChemistrySection B • Numericalcoordination compounds
+4 / -1
The number of paramagnetic species from the following is _____________.
[Ni(CN)4]2−,[Ni(CO)4],[NiCl4]2−
[Fe(CN)6]4−,[Cu(NH3)4]2+
[Fe(CN)6]3−and[Fe(H2O)6]2+
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
Species [Ni(CN)4]2−[Ni(CO)4][NiCl]4]2−[FeCN)6]4−[Fe(CN)6]3−Fe(H2O)6]2+[Cu(NH3)4]2+ Magnetic property Diamagnetic Diamagnetic Paramagnetic Diamagnetic Paramagnetic Paramagnetic Paramagnetic
Q 61MathematicsSection A • MCQdefinite integration
+4 / -1
The minimum value of the function f(x)=0∫2e∣x−t∣dt is :
A
2
B
2(e−1)
C
e(e−1)
D
2e−1
Official Correct Answer:Option (B)
Step-by-Step Solution:
f(x)=∫02e∣x−t∣dt
For x>2
f(x)=∫02ex−tdt=ex(1−e−2)
For x<0
f(x)=∫02et−xdt=e−x(e2−1)
For x∈[0,2]
f(x)=∫0xex−tdt+∫x2et−xdt=e2−x+ex−2
For x>2
f(x)∣min=e2−1
For x<0
f(x)∣min=e2−1
For x∈[0,2]
f(x)∣min=2(e−1)
Q 62MathematicsSection B • Numerical3d geometry
+4 / -1
Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z−7=0=2x+3y+z−2 be ax+by+cz=65. Then the distance of the point (a, b, c) from the plane 2x+2y−z+16=0 is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 9
Official Correct Answer:9
Step-by-Step Solution:
Let the equation of the plane is
(x−2y−z−5)+λ(x+y+3z−5)=0
∵ it's parallel to the line
x+y+2z−7=0=2x+3y+z−2
So, vector along the line i^12j^13k^21
=−5i^+3j^+k^
∵ Plane is parallel to line
∴−5(1+λ)+3(−2+λ)+1(−1+3λ)=0
λ=12
So, by (i)
13x+10y+35z=65
∴a=13,b=10,c=35
and d=926+20−35+16=9
Q 63MathematicsSection A • MCQparabola
+4 / -1
The distance of the point (6,−22) from the common tangent y=mx+c,m>0, of the curves x=2y2 and x=1+y2 is :
A
31
B
5
C
314
D
53
Official Correct Answer:Option (B)
Step-by-Step Solution:
y2=2x⇒ tangent y=mx+8m1y2=x−1⇒ tangent y=m(x−1)+4m1 For common tangent 8m1=−m+4m1⇒1=−8m2+2∵m>0⇒m=221⇒ Common tangent is y=22x+221⇒x−22y+1=0
Distance of point (6,−22) from common tangent =5
Q 64MathematicsSection A • MCQindefinite integration
+4 / -1
Let f(x)=∫(x2+1)(x2+3)2xdx. If f(3)=21(loge5−loge6), then f(4) is equal to
A
loge19−loge20
B
loge17−loge18
C
21(loge19−loge17)
D
21(loge17−loge19)
Official Correct Answer:Option (D)
Step-by-Step Solution:
f(x)=∫(x2+1)(x2+3)2xdx
Put x2=t⇒2xdx=dtf(x)=∫(t+1)(t+3)dt=∫(t+2)2−1dt=21loget+3t+1+Cf(x)=21loge(x2+3x2+1)+C⇒f(3)=21loge(1210)+C∵f(3)+21(loge5−loge6)⇒C=0f(x)=21loge(x2+3x2+1)⇒f(4)=21(loge17−loge19)
Q 65MathematicsSection B • Numericalsets and relations
+4 / -1
Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 43
Official Correct Answer:43
Step-by-Step Solution:
Elements of the type 3k=3
Elements of the type 3k+1=1,7,9
Elements of the type 3k+2=2,5,11
Subsets containing one element S1=1
Subsets containing two elements
S2=3C1×3C1=9
Subsets containing three elements
S3=3C1×3C1+1+1=11
Subsets containing four elements
<br><br>S4=3C3+3C3+3C2×3C2=11
Subsets containing five elements
S5=3C2×3C2×1=9
Subsets containing six elements S6=1
Subsets containing seven elements S7=1
⇒ sum =43
Q 66MathematicsSection A • MCQprobability
+4 / -1
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S={x∈Z:x(66−x)≥95M} and the event A={x∈S:xisamultipleof3}. Then P(A) is equal to :
The vector a=−i+2j+k is rotated through a right angle, passing through the y-axis in its way and the resulting vector is b. Then the projection of 3a+2b on c=5i+4j+3k is :
A
6
B
23
C
1
D
32
Official Correct Answer:Option (D)
Step-by-Step Solution:
First, we write b as a linear combination of a and j since b is a rotation of a about the y-axis.
b=λa+μj^=λ(−i^+2j^+k^)+μj^=−λi^+(2λ+μj^)+λk^
b is orthogonal to a due to the right angle rotation, so b⋅a=0.
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :
The vertices of a hyperbola H are (± 6, 0) and its eccentricity is 25. Let N be the normal to H at a point in the first quadrant and parallel to the line 2x+y=22. If d is the length of the line segment of N between H and the y-axis then d2 is equal to _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 216
Official Correct Answer:216
Step-by-Step Solution:
H:36x2−9y2=1
equation of normal is 6xcosθ+3ycotθ=45
slope =−2sinθ=−2⇒θ=4π
Equation of normal is 2x+y=15
P:(asecθ,btanθ)
⇒P(62,3) and K(0,15)
d2=216
Q 71MathematicsSection A • MCQcomplex numbers
+4 / -1
Let z1=2+3i and z2=3+4i. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2} represents a
A
hyperbola with the length of the transverse axis 7
B
hyperbola with eccentricity 2
C
straight line with the sum of its intercepts on the coordinate axes equals −18
D
straight line with the sum of its intercepts on the coordinate axes equals 14
Official Correct Answer:Option (D)
Step-by-Step Solution:
∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2
⇒(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1
⇒−4x+4+9−6y−9+6x−16+8y=2
⇒2x+2y=14
⇒x+y=7
Q 72MathematicsSection A • MCQmatrices and determinants
+4 / -1
Let S1 and S2 be respectively the sets of all a∈R−{0} for which the system of linear equations
ax+2ay−3az=1
(2a+1)x+(2a+3)y+(a+1)z=2
(3a+5)x+(a+5)y+(a+2)z=3
has unique solution and infinitely many solutions. Then
A
n(S1)=2 and S2 is an infinite set
B
S1=Φ and S2=R−{0}
C
S1=R−{0} and S2=Φ
D
S1 is an infinite set and n(S2) = 2
Official Correct Answer:Option (C)
Step-by-Step Solution:
Given system of equations
ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3 Let A=a2a+13a+52a2a+3a+5−3aa+1a+2=a12a+13a+501−2a−5a−507a+410a+17=a(15a2+31a+37) Now A=0⇒a=0
So, S1=R−{0} and at a=0
System has infinite solution but a∈R−{0}
∴S2=Φ
Q 73MathematicsSection A • MCQmathematical reasoning
+4 / -1
The statement (p∧(∼q))⇒(p⇒(∼q)) is
A
a tautology
B
equivalent to (∼p)∨(∼q)
C
a contradiction
D
p∨q
Official Correct Answer:Option (A)
Step-by-Step Solution:
Making truth table (Let (p∧∼q)⇒(p⇒∼q)=E )
p
q
∼p
∼q
p∧∼q
p⇒∼q
E
T
T
F
F
F
F
T
T
F
F
T
T
T
T
F
T
T
F
F
T
T
F
F
T
T
F
T
T
∴E is a tautology
Q 74MathematicsSection B • Numericalinverse trigonometric functions
+4 / -1
If the sum of all the solutions of tan−1(1−x22x)+cot−1(2x1−x2)=3π,−1<x<1,x=0, is α−34, then α is equal to _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
Case-I
−1<x<0
tan−1(1−x22x)+π+tan−1(1−x22x)=3π
tan−11−x22x=3−π
2tan−1x=3−π
tan−1x=6−π
x=3−1
Case-II
0<x<1
tan−11−x22x+tan−11−x22x=3π
tan−11−x22x=6π2tan−1x=6πtan−1x=12πx=2−3 Sum =3−1+2−3=2−34⇒α=2
Q 75MathematicsSection B • Numericalfunctions
+4 / -1
For some a, b, c ∈N, let f(x)=ax−3 and g(x)=xb+c,x∈R. If (fog)−1(x)=(2x−7)1/3, then (fog)(ac)+(gof)(b) is equal to ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2039
Official Correct Answer:2039
Step-by-Step Solution:
f(x)=ax−3
g(x)=xb+c
(fog)−1=(2x−7)31
(fog)−1(x)=(ax+3−ca)b1=(2x−7)31
⇒a=2,b=3,c=5
fog(ac)+gof(b)
∵f(x)=2x−3
g(x)=x3+5
fog(10)+gof(3)
=2007+32
=2039
Q 76MathematicsSection B • Numericalpermutations and combinations
+4 / -1
Let x and y be distinct integers where 1≤x≤25 and 1≤y≤25. Then, the number of ways of choosing x and y, such that x+y is divisible by 5, is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 120
Official Correct Answer:120
Step-by-Step Solution:
Let x+y=5λ
Possible cases are
x5λ(5,10,15,20,25)5λ+1(1,6,11,16,21)5λ+2(2,7,12,17,22)5λ+3(3,8,13,18,23)5λ+4(4,9,14,19,24)y5λ(5,10,15,20,25)5λ+4(4,9,14,19,24)5λ+3(3,8,13,18,23)5λ+2(2,7,12,17,22)5λ+1(1,6,11,16,21) Number of ways 2025252525
Total number of ways =20+25+25+25+25=120
Note : In first case total number of ways = 20 as in the question given that the chosen x and y are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.
Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.
∴ Total number of ways in the first case = 5 × 4 = 20.
Q 77MathematicsSection A • MCQcircle
+4 / -1
The points of intersection of the line ax+by=0,(a=b) and the circle x2+y2−2x=0 are A(α,0) and B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0 is :
A
x2+y2+5x+5y+12=0
B
x2+y2+3x+5y+8=0
C
x2+y2−5x−5y+12=0
D
x2+y2+3x+3y+4=0
Official Correct Answer:Option (A)
Step-by-Step Solution:
As A and B satisfy both line and circle we have α=0⇒A(0,0) and β=1 i.e. B(1,1)
Centre of circle as AB diameter is (21,21) and radius =21
∴ For image of (21;21) in x+y+z we get 1x−21=1y−21=2−2(3)
⇒ Image (−25,−25)
∴ Equation of required circle
(x+25)2+(y+25)2=21
⇒x2+y2+5x+5y+450−21=0
⇒x2+y2+5x+5y+12=0
Q 78MathematicsSection A • MCQapplication of derivatives
+4 / -1
Let x=2 be a local minima of the function f(x)=2x4−18x2+8x+12,x∈(−4,4). If M is local maximum value of the function f in (−4,4), then M =
Q 79MathematicsSection B • Numericalmatrices and determinants
+4 / -1
Let A1,A2,A3 be the three A.P. with the same common difference d and having their first terms as A,A+1,A+2, respectively. Let a, b, c be the 7th,9th,17th terms of A1,A2,A3, respective such that \left| {\matrix{
a & 7 & 1 \cr
{2b} & {17} & 1 \cr
c & {17} & 1 \cr
} } \right| + 70 = 0.
If a=29, then the sum of first 20 terms of an AP whose first term is c−a−b and common difference is 12d, is equal to ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 495
Q 82MathematicsSection B • Numericalarea under the curves
+4 / -1
If the area enclosed by the parabolas P1:2y=5x2 and P2:x2−y+6=0 is equal to the area enclosed by P1 and y=αx,α>0, then α3 is equal to ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 600
Official Correct Answer:600
Step-by-Step Solution:
x2+6=25x2⇒x=±2
Area between P1 and P2 [Say A1]=−2∫2(x2+6)−25x2dx=20∫2(6−23x2)dx=2[6x−2x3]02=16
ax=25x2⇒x=0,52a Area between P1 and y=ax [Say A2 ] =0∫52αax−25x2dx=2ax2−65x3]052a:752a3A1=A2⇒752a3=16a3=600
Q 83MathematicsSection A • MCQdifferential equations
+4 / -1
Let y=y(x) be the solution curve of the differential equation dxdy=xy(1+xy2(1+logex)),x>0,y(1)=3. Then 9y2(x) is equal to :
A
5−2x3(2+logex3)x2
B
3x3(1+logex2)−2x2
C
7−3x3(2+logex2)x2
D
2x3(2+logex3)−3x2
Official Correct Answer:Option (A)
Step-by-Step Solution:
dxdy−xy=y3(1+logex)y31dxdy−xy21=1+logex Let −y21=t⇒y32dxdy=dxdt∴dxdt+x2t=2(1+logex) I.F. =e∫x2dx=x2y2−x2=32((1+logex)x3−3x3)+Cy(1)=39y2=5−2x3(2+logex3)C
Q 84MathematicsSection B • Numericalbinomial theorem
+4 / -1
The constant term in the expansion of (2x+x71+3x2)5 is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1080
====5C4(x8(3x+2))45C4 coefficient of x3 in (2+3x)45C4×4C3(2)1(3)35×4×2×271080
Q 85MathematicsSection A • MCQ3d geometry
+4 / -1
Consider the lines L1 and L2 given by
L1:2x−1=1y−3=2z−2
L2:1x−2=2y−2=3z−3.
A line L3 having direction ratios 1, −1, −2, intersects L1 and L2 at the points P and Q respectively. Then the length of line segment PQ is
A
43
B
26
C
4
D
32
Official Correct Answer:Option (B)
Step-by-Step Solution:
Let,
P≡(2λ+1,λ+3,2λ+2) and Q(μ+2,2μ+2, 3μ+3) d.r’s of PQ≡<2λ−μ−1,λ−2μ+1,2λ−3μ−1>∴12λ−μ−1−−1λ−2μ−1=−22λ−3μ−1∴−2λ+μ+1=λ−2μ+1 and −2λ+4μ−2=−2λ+3μ+1⇒3λ−3μ=0 and μ=3∴λ=±3 and μ=3∴P≡(7,6,8) and Q(5,8,12)∴∣PO∣=22+22+42=24=26
Q 86MathematicsSection B • Numericallogarithms
+4 / -1
Let S={α:log2(92α−4+13)−log2(25.32α−4+1)=2}. Then the maximum value of β for which the equation x2−2(α∈s∑α)2x+α∈s∑(α+1)2β=0 has real roots, is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 25
Official Correct Answer:25
Step-by-Step Solution:
log2(92α−4+13)−log2(25⋅32α−4+1)=2⇒2532α−4+192α−4+13=4⇒α=2 or 3α∈S∑α=5 and α∈S∑(α+1)2=25⇒x2−50x+25β=0 has real roots ⇒β≤25⇒βmax=25
Q 87MathematicsSection A • MCQdifferentiation
+4 / -1
Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′ at x=−1 is equal to
The distance of the point P(4, 6, −2) from the line passing through the point (−3, 2, 3) and parallel to a line with direction ratios 3, 3, −1 is equal to :
A
3
B
14
C
6
D
23
Official Correct Answer:Option (B)
Step-by-Step Solution:
AP=7i^+4j^−5k^⇒∣AP∣=49+16+25=90AN
= projection of AP on b=AP⋅b=1921+12+5=1938
(PN)2=(AP)2−(AN)2=90−76=14⇒PN=14
Q 90MathematicsSection A • MCQapplication of derivatives
+4 / -1
Let f:(0,1)→R be a function defined f(x)=1−e−x1, and g(x)=(f(−x)−f(x)). Consider two statements