Official NTA Question PaperFull Solved Paper2021Morning Shift
JEE Main 2021 31st August 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.
The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :
A
V=21r4G(M1+M2)
B
V=r4G(M1+M2)
C
V=21r2G(M1+M2)
D
V=r2G(M1+M2)
Official Correct Answer:Option (B)
Step-by-Step Solution:
21mV2−r/2GM1m−r/2GM2m=0
21mV2=r2Gm(M1+M2)
V=r4G(M1+M2)
Option (b)
Q 2PhysicsSection A • MCQproperties of matter
+4 / -1
A uniform heavy rod of weight 10 kg ms−2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
A
2 × 10−9 m
B
5 × 10−8 m
C
4 × 10−8 m
D
5 × 10−10 m
Official Correct Answer:Option (D)
Step-by-Step Solution:
We know,
Δl=2AYWL
Δl=2×510×1×100×10−4×2×1011
Δl=21×10−9=5×10−10 m
Option (d)
Q 3PhysicsSection A • MCQheat and thermodynamics
+4 / -1
For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation dvdp=−ap. If p = p0 at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)
A
aeRp0
B
eRap0
C
infinity
D
0∘C
Official Correct Answer:Option (A)
Step-by-Step Solution:
p0∫pPdp=−a0∫vdv
ln(p0p)=−av
p=p0e−av
For temperature maximum p-v product should be maximum
T=nRpv=Rp0ve−av
dvdT=0⇒Rp0{e−av+ve−av(−a)} = 0
Rp0e−av{1−av}=0
v=a1,∞
T=Raep01=Raep0
at v = ∞
T = 0
Option (a)
Q 4PhysicsSection A • MCQray optics
+4 / -1
Two plane mirrors M1 and M2 are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M1 and M2 respectively. The shortest distance between the images thus formed is : (Take 5 = 2.3)
A
3a
B
4.6a
C
2.3a
D
210a
Official Correct Answer:Option (B)
Step-by-Step Solution:
Shortest distance is 2a between I1 & I3
But answer given is for I1 & I2
= (4a)2+(2a)2
= a20
= 4.47 a
Option (b)
Q 5PhysicsSection A • MCQelectrostatics
+4 / -1
Two particles A and B having charges 20μC and −5μC respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?
A
At 5 cm from 20 μC on the left side of system
B
At 5 cm from −5 μC on the right side
C
At 1.25 cm from −5 μC between two charges
D
At midpoint between two charges
Official Correct Answer:Option (B)
Step-by-Step Solution:
Null point is possible only right side of −5 μC
EN=+x2k(−5μC)+(5+x)2k(20μC)=0
x = 5 cm
∴ Option (b) is correct.
Q 6PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A reversible engine has an efficiency of 41. If the temperature of the sink is reduced by 58∘C, its efficiency becomes double. Calculate the temperature of the sink :
A
174∘C
B
280∘C
C
180.4∘C
D
382∘C
Official Correct Answer:Option (A)
Step-by-Step Solution:
T2 = sink temperature
η=1−T1T2
41=1−T1T2
T1T2=43 .... (i)
21=1−T1T2−58
T1T2=T158=21
43=T158+21
41=T158⇒T1=232
T2=43×232
T2=174 K
Q 7PhysicsSection A • MCQatoms and nuclei
+4 / -1
A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B versus time is :
(Assume that at t = 0, there are no B atoms in the sample)
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
A → B → C (stable)
Initially no. of atoms of B = 0 at t = 0, no. of atoms of B will starts increasing & reaches maximum value when rate of decay of B = rate of formation of B.
After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions, so best possible graph is (d).
Q 8PhysicsSection B • Numericalproperties of matter
+4 / -1
When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%.
(The bulk modulus of rubber = 9.8 × 108 Nm−2, Density of sea water = 103 kgm−3, g = 9.8 m/s2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 500
Official Correct Answer:500
Step-by-Step Solution:
B=−(VΔV)ΔP=−(VΔV)ρgh
−ρgBVΔV=h
100×103×9.89.8×108×0.5=h
h = 500
Q 9PhysicsSection B • Numericalcurrent electricity
+4 / -1
The voltage drop across 15Ω resistance in the given figure will be ______________ V.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 6
Official Correct Answer:6
Step-by-Step Solution:
⇒ effective circuit diagram will be
Point drop across 6Ω = 1 × 6 = 6 = VAB
⇒ Hence point drop across 15Ω = 6 volt = VAB
Q 10PhysicsSection B • Numericalwork power and energy
+4 / -1
A block moving horizontally on a smooth surface with a speed of 40 ms−1 splits into two equal parts. If one of the parts moves at 60 ms−1 in the same direction, then the fractional change in the kinetic energy will be x : 4 where x = ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
Pi = Pf
m × 40 = 2m× v + 2m× 60
40 = 2v + 30
⇒ v = 20
(K. E.)I = 21m × (40)2 = 800 m
(K. E.)f = 212m . (20)2 + 21 . 2m (60)2 = 1000 m
| Δ K. E. | = | 1000m − 800 m | = 200 m
(K.E.)iΔK.E.=800m200m=41=4x
x = 1
Q 11PhysicsSection A • MCQcurrent electricity
+4 / -1
Consider a galvanometer shunted with 5Ω resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?
A
300 Ω
B
344 Ω
C
245 Ω
D
226 Ω
Official Correct Answer:Option (C)
Step-by-Step Solution:
0.02i Rg = 0.98i × 5
Rg = 245 Ω
Option (c)
Q 12PhysicsSection A • MCQunits and measurements
+4 / -1
Match List - I with List - II.
List - I
List - II
(a)
Torque
(i)
MLT−1
(b)
Impulse
(ii)
MT−2
(c)
Tension
(iii)
ML2T−2
(d)
Surface Tension
(iv)
MLT−2
Choose the most appropriate answer from the option given below :
A
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
C
(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
D
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Official Correct Answer:Option (A)
Step-by-Step Solution:
torque τ→ ML2T−2 (iii)
Impulse I ⇒ MLT−1 (i)
Tension force ⇒ MLT−2 (iv)
Surface tension ⇒ MT−2 (ii)
Option (a)
Q 13PhysicsSection B • Numericallaws of motion
+4 / -1
A car is moving on a plane inclined at 30∘ to the horizontal with an acceleration of 10 ms−2 parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ______________. (Take g = 10 ms−2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 30
Official Correct Answer:30
Step-by-Step Solution:
Given,
Angle of inclination, θ = 30∘
Acceleration, a = 10 ms−2
Acceleration due to gravity, g = 10 ms−2
According to the question the car and bob is as shown below,
Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.
In equilibrium, ΣFx = 0 and ΣFy = 0
⇒ F' cos30∘ = T sinα
sinαmacos30∘=T ...... (i)
where, m is the mass of the bob.
F′sin30∘+mg=Tcosα
⇒masin30∘+mg=sinαmacos30∘(cosα) [∵ using Eq. (i)]
⇒asin30∘+g=sinαacos30∘(cosα)
10×21+10=210×3cotα
⇒21+1=23cotα⇒23=23cotα
or, cotα=3
⇒α=30∘
Q 14PhysicsSection A • MCQelectronic devices
+4 / -1
In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be :
A
1, 0, 1, 0
B
0, 1, 0, 1
C
1, 1, 1, 0
D
0, 0, 1, 1
Official Correct Answer:Option (C)
Step-by-Step Solution:
Y=(A.B).(A+B)
Y)(0,0)=1
Y)(0,1)=1
Y)(1,0)=1
Y)(1,1)=0
Option (c) is correct.
Q 15PhysicsSection B • Numericalcommunication systems
+4 / -1
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 64
Official Correct Answer:64
Step-by-Step Solution:
hT = hR = 160 ........ (i)
d=2RhT+2RhR
d=2R[hT+hR]
d=2R[x+160−x]
dxd(d)=0
2x1+2160−x1(−1)=0
x1=160−x1
x = 80 m
dmax=2×6400[100080+100020]
=1010802×280
=8×2×2×22=64 km
Q 16PhysicsSection A • MCQrotational motion
+4 / -1
Angular momentum of a single particle moving with constant speed along circular path :
A
changes in magnitude but remains same in the direction
B
remains same in magnitude and direction
C
remains same in magnitude but changes in the direction
D
is zero
Official Correct Answer:Option (B)
Step-by-Step Solution:
L = mvr
And direction will be upward & remain constant.
Option (b)
Q 17PhysicsSection B • Numericalsimple harmonic motion
+4 / -1
A particle of mass 1 kg is hanging from a spring of force constant 100 Nm−1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is xT. The value of x is _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 8
Official Correct Answer:8
Step-by-Step Solution:
KE = PE
y=2A=Asinωt
t=8T=xT
x = 8
Q 18PhysicsSection B • Numericalelectromagnetic waves
+4 / -1
The electric field in an electromagnetic wave is given by E = (50 NC−1) sinω (t − x/c)
The energy contained in a cylinder of volume V is 5.5 × 10−12 J. The value of V is _____________ cm3. (given ∈0 = 8.8 × 10−12C2N−1m−2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 500
Official Correct Answer:500
Step-by-Step Solution:
E=50sin(ωt−cω.x)
Energy density = 21∈0E02
Energy of volume V=21∈0E02.V=5.5×10−12
218.8×10−12×2500V=5.5×10−12
V=2500×8.85.5×2=.0005m3
= .0005 × 106 (c.m)3
= 500 (c.m)3
Q 19PhysicsSection A • MCQmotion
+4 / -1
A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
A
h22ghv2+1
B
2ghv2+h2
C
g2v2h+h2
D
v22gh+h2
Official Correct Answer:Option (C)
Step-by-Step Solution:
R=g2h.v
D=R2+h2
=(g2h.v)2+h2
D=g2hv2+h2
Option (c) is correct.
Q 20PhysicsSection A • MCQelectronic devices
+4 / -1
Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one :
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
When Vi > 3 volt, VR > 0
Because diode will be in forward biased state
When Vi≤ 3 volt ; VR = 0
Because diode will be in reverse biased state.
Q 21PhysicsSection A • MCQray optics
+4 / -1
An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?
A
1, ∞
B
Very high, ∞
C
21, 2f
D
41, 4f
Official Correct Answer:Option (C)
Step-by-Step Solution:
U = −f
V1−U1=−f1⇒V1=−f2
V=2−f
m=UV=21
distance = 2f
Option (c)
Q 22PhysicsSection A • MCQdual nature of radiation
+4 / -1
A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
A
Kp < Ke and Pp = Pe
B
Kp = Ke and Pp = Pe
C
Kp < Ke an Pp < Pe
D
Kp > Ke and Pp = Pe
Official Correct Answer:Option (A)
Step-by-Step Solution:
λp=Pph
λe=Peh
∵λp=λe
⇒Pp=Pe
(K)p=2mpPp2
(K)e=2mePe2
Kp < Ke as mp > me
Option (a)
Q 23PhysicsSection B • Numericalcapacitor
+4 / -1
A capacitor of 50 μF is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ______________μC.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 100
Official Correct Answer:100
Step-by-Step Solution:
Potential Difference across each resistor = 2V
q = CV
= 50 × 10−6× 2 = 100 × 10−6 = 100 μC
Q 24PhysicsSection A • MCQalternating current
+4 / -1
In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL = R = XC. Impedance of this circuit is :
A
2R2
B
Zero
C
R
D
R2
Official Correct Answer:Option (C)
Step-by-Step Solution:
Z=(XL−XC)2+R2=R∵ XL = XC
Option (c)
Q 25PhysicsSection B • Numericalwaves
+4 / -1
A wire having a linear mass density 9.0 × 10−4 kg/m is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ____________ m.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 10
Official Correct Answer:10
Step-by-Step Solution:
μ=9.0×10−4mkg
T = 900 N
V=μT=9×10−4900=1000 m/s
f1 = 500 Hz
f = 550
2lnV=500 .... (i)
2l(n+1)V=500 .... (ii)
(ii) (i) 2lV=50
l=2×501000=10
Q 26PhysicsSection A • MCQmagnetics
+4 / -1
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:
A
2(b−a)μ0INloge(ab)
B
8μ0I[a−ba+b]
C
4(a−b)μ0I[a1−b1]
D
8μ0I(a+ba−b)
Official Correct Answer:Option (A)
Step-by-Step Solution:
No. of turns in dx width = b−aNdx
∫dB=a∫b(b−aN)dx2xμ0I
B=2(b−a)Nμ0iln(ab)
Option (a)
Q 27PhysicsSection A • MCQelectromagnetic induction
+4 / -1
A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
A
4πμ082ba2
B
4πμ0a82
C
4πμ082ab2
D
4πμ0b82
Official Correct Answer:Option (A)
Step-by-Step Solution:
B=[4πμ0b/2I×2sin45]×4
ϕ=22πμ0bI×a2
∴M=Iϕ=πb22μ0a2=4πμ082ba2
Option (a)
Q 28PhysicsSection B • Numericalcurrent electricity
+4 / -1
A square shaped wire with resistance of each side 3Ω is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of Ω will be ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 3
Official Correct Answer:3
Step-by-Step Solution:
Req = 3Ω
Q 29PhysicsSection A • MCQunits and measurements
+4 / -1
Which of the following equations is dimensionally incorrect?
Where t = time, h = height, s = surface tension, θ = angle, ρ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque, ∈ = permittivity, E = electric field, J = current density, L = length.
A body of mass M moving at speed V0 collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles θ1 and θ2 with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles θ1 and θ2 will be equal, is :
A
4
B
1
C
3
D
2
Official Correct Answer:Option (C)
Step-by-Step Solution:
Given θ1 = θ2 = θ
from momentum conservation
in x-direction MV0 = MV1 cosθ + mV2 cosθ
in y-direction 0 = MV1 sinθ− mV2 sinθ
Solving above equations
V2=mMV1, V0 = 2V1 cosθ
From energy conservation
21MV02=21MV12+21MV22
Substituting value of V2 & V0, we will get
mM+1=4cos2θ≤4
mM≤3
Option (c)
Q 31ChemistrySection A • MCQpolymers
+4 / -1
Monomer of Novolac is :
A
3-Hydroxybutanoic acid
B
phenol and melamine
C
o-Hydroxymethylphenol
D
1, 3-Butadiene and styrene
Official Correct Answer:Option (C)
Step-by-Step Solution:
Q 32ChemistrySection A • MCQenvironmental chemistry
+4 / -1
BOD values (in ppm) for clean water (A) and polluted water (B) are expected respectively :
A
A > 50, B < 27
B
A > 25, B < 17
C
A < 5, B > 17
D
A > 15, B > 47
Official Correct Answer:Option (C)
Step-by-Step Solution:
BOD values of clean water (A) is less than 5 ppm
So, A < 5
BOD values of polluted water (B) is greater than 17 ppm
So, B > 17
So, Ans. is (c)
Q 33ChemistrySection B • Numericalionic equilibrium
+4 / -1
A3B2 is a sparingly soluble salt of molar mass M (g mol−1) and solubility x g L−1. The solubility product satisfies Ksp=a(Mx)5. The value of a is _____________. (Integer answer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 108
Official Correct Answer:108
Step-by-Step Solution:
KSP = (3s)3 (2s)2
KSP = 108 s5 & s = (x/M)
KSP = 108(Mx)5
given Ksp=a(Mx)5
comparing a = 108
Q 34ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
The correct order of reactivity of the given chlorides with acetate in acetic acid is :
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Acetate in acetic acid with R-X cause SN1
Allylic halides undergo faster SN1
∴ Correct order is
Q 35ChemistrySection B • Numericalcompounds containing nitrogen
+4 / -1
The total number of reagents from those given blow, that can convert nitrobenzene into aniline is ___________. (Integer answer)
I. Sn − HCl
II. Sn − NH4OH
III. Fe − HCl
IV. Zn − HCl
V. H2− Pd
VI. H2− Raney Nickel
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
(i) Sn + HCl
(ii) Fe + HCl
(iii) Zn + HCl
(iv) H2− Pd
(v) H2 (Raney Ni)
Q 36ChemistrySection A • MCQhydrogen
+4 / -1
Given below are two statements :
Statement I : The process of producing syn-gas is called gasification of coal.
Statement II : The composition of syn-gas is CO + CO2 + H2 (1 : 1 : 1)
In the light of the above statements, choose the most appropriate answer from the options given below :
A
Statement I is false but Statement II is true
B
Statement I is true but Statement II is false
C
Both Statement I and Statement II are false
D
Both Statement I and Statement II are true
Official Correct Answer:Option (B)
Step-by-Step Solution:
The process of producing syn-gas from coal is called gasification of coal.
Syn-gas having composition of CO & H2 in 1 : 1
Q 37ChemistrySection A • MCQisolation of elements
+4 / -1
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Aluminium is extracted from bauxite by the electrolysis of molten mixture of Al2O3 with cryolite.
Reason (R) : The oxidation state of Al in cryolite is +3.
In the light of the above statements, choose the most appropriate answer from the options given below :
A
(A) is true but (R) is false.
B
(A) is false but (R) is true.
C
Both (A) and (R) are correct and (R) is the correct explanation of (A)
D
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Official Correct Answer:Option (D)
Step-by-Step Solution:
(A) Aluminium is reactive metal so Aluminium is extracted by electrolysis of Alumina with molten mixture of Cryolite
(B) Cryolite, Na3AlF6
Here, Al is in +3 O.S.
So, answer is (d).
Q 38ChemistrySection B • Numericalchemical kinetics and nuclear chemistry
+4 / -1
For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ____________. (Integer answer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
k=t2.303loga−xa
t50%2.303log100−50100=t75%2.303log100−75100
⇒t75%=2t50%
Q 39ChemistrySection 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) : Treatment of bromine water with propene yields 1-bromopropan-2-ol.
Reason (R) : Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol.
In the light of the above statements, choose the most appropriate 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 false but (R) is true
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 (C)
Step-by-Step Solution:
Its IUPAC name 1-bromopropan-2-ol
A and R are true and (R) is the correct explanation of (A).
Q 40ChemistrySection A • MCQsolutions
+4 / -1
Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?
A
hydrazine
B
glucose
C
glycine
D
KHSO4
Official Correct Answer:Option (D)
Step-by-Step Solution:
ΔTf∝ i × m
greater the value of i, greater will be the ΔTf
value.
Van't Hoff factor is highest for KHSO4
∴ colligative property (ΔTf) will be highest for KHSO4
Q 41ChemistrySection B • Numericalelectrochemistry
The value of Ecell0 is 4.315 V at 25∘C. If ΔH∘ = −825.2 kJ mol−1, the standard entropy change ΔS∘ in J K−1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol−1]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 25
Official Correct Answer:25
Step-by-Step Solution:
ΔG∘=−nFE∘=ΔH∘−TΔS∘
∴ΔS∘=TΔH∘+nFE∘
=298(−825.2×103)+(2×96487×4.315)
=298−825.2×103+832.682×103
=2987.483×103=25.11 JK−1 mol−1
∴ Nearest integer answer is 25.
Q 42ChemistrySection A • MCQalcohols phenols and ethers
+4 / -1
The major product formed in the following reaction is :
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 43ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L−1 is x × 10−2. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 20
Official Correct Answer:20
Step-by-Step Solution:
[H2C2O4.2H2O]=V(L)weight/Mw
⇒x×10−2=250/10006.3/126
x=20
Q 44ChemistrySection A • MCQsurface chemistry
+4 / -1
Select the graph that correctly describes the adsorption isotherms at two temperatures T1, and T2 (T1 > T2) for a gas :
(x - mass of the gas absorbed ; m - mass of adsorbent ; P - pressure)
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
According to Freundlich adsorption isotherm, in the median range of pressure
mx∝Pn1
⇒mx = kPn1
On increasing temperature mx decreases.
∵ Adsorption is generally exothermic in nature. Hence, correct option is (A).
Q 45ChemistrySection A • MCQs block elements
+4 / -1
The major component/ingredient of Portland Cement is :
A
tricalcium aluminate
B
tricalcium silicate
C
dicalcium aluminate
D
dicalcium silicate
Official Correct Answer:Option (B)
Step-by-Step Solution:
Major component of Portland cement is "Tricalcium silicate (51%, 3CaO.SiO2)
Q 46ChemistrySection A • MCQd and f block elements
+4 / -1
In the structure of the dichromate ion, there is a :
A
linear symmetrical Cr-O-Cr bond.
B
non-linear symmetrical Cr-O-Cr bond.
C
linear unsymmetrical Cr-O-Cr bond.
D
non-linear unsymmetrical Cr-O-Cr bond.
Official Correct Answer:Option (B)
Step-by-Step Solution:
Dichromate ion contain non-linear symmetrical Cr-O-Cr bond.
Q 47ChemistrySection A • MCQcoordination compounds
+4 / -1
The denticity of an organic ligand, biuret is :
A
2
B
4
C
3
D
6
Official Correct Answer:Option (A)
Step-by-Step Solution:
Biuret ligand is
It forms complexes like
The denticity of organic ligand is 2.
Q 48ChemistrySection A • MCQcompounds containing nitrogen
+4 / -1
The major products A and B in the following set of reactions are :
A
B
C
D
Official Correct Answer:Option (B)
Step-by-Step Solution:
Q 49ChemistrySection B • Numericalp block elements
+4 / -1
Consider the sulphides HgS, PbS, CuS, Sb2S3, As2S3 and CdS. Number of these sulphides soluble in 50% HNO3 is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
Pbs, CuS, As2S3, CdS are soluble in 50% HNO3.
HgS, Sb2S3 are insoluble in 50% HNO3
So, answer is 4.
Q 50ChemistrySection A • MCQd and f block elements
+4 / -1
Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature? (Given Z for Nd = 60, Yb = 70, La = 57, Ce = 58)
A
Nd
B
Yb
C
La
D
Ce
Official Correct Answer:Option (B)
Step-by-Step Solution:
Yb (70) = 4f14 6s2
Yb+2 = 4f14 6s0
∵ All the electrons are paired hence Yb+2 is
diamagnetic.
Q 51ChemistrySection B • Numericalchemical bonding and molecular structure
+4 / -1
The number of hydrogen bonded water molecule(s) associated with stoichiometry CuSO4.5H2O is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
One hydrogen bonded H2O molecule
Q 52ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
The structure of product C, formed by the following sequence of reactions is :
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 53ChemistrySection A • MCQbasics of organic chemistry
+4 / -1
Choose the correct name for compound given below :
A
(4E)-5-Bromo-hex-4-en-2-yne
B
(2E)-2-Bromo-hex-4-yn-2-ene
C
(2E)-2-Bromo-hex-2-en-4-yne
D
(4E)-5-Bromo-hex-2-en-4-yne
Official Correct Answer:Option (C)
Step-by-Step Solution:
First prioritise the groups according to CIP rule
It is (2E)-2-Bromo-hex-2-en-4-yne
Q 54ChemistrySection B • Numericalatomic structure
+4 / -1
Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with ml = 0. The value of x is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 7
Official Correct Answer:7
Step-by-Step Solution:
Completely filled orbital with ml = 0 are
= 1 + 1 + 1 + 1 + 1 + 1 + 1
= 7
So, answer is 7.
Q 55ChemistrySection A • MCQbiomolecules
+4 / -1
Which one of the following compounds contains β-C1-C4 glycosidic linkage?
A
Lactose
B
Sucrose
C
Maltose
D
Amylose
Official Correct Answer:Option (A)
Step-by-Step Solution:
In Lactose it is β C1 - C4 glycosidic linkage.
In Maltose, Amylose α C1 - C4 glycosidic linkage is present.
Q 56ChemistrySection B • Numericalchemical kinetics and nuclear chemistry
+4 / -1
According to the following figure, the magnitude of the enthalpy change of the reaction
A + B → M + N in kJ mol− is equal to ___________. (Integer answer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 45
Official Correct Answer:45
Step-by-Step Solution:
ΔH = EProduct− EReactant
= 15 − (15 + 45)
= −45 KJ/mol
| ΔH | = 45 KJ/mol
Q 57ChemistrySection A • MCQperiodic classification of elements
+4 / -1
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Metallic character decreases and non-metallic character increases on moving from left to right in a period.
Reason (R) : It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period.
In the light of the above statements, choose the most appropriate answer from the options given below :
A
(A) is false but (R) is true.
B
(A) is true but (R) is false
C
Both (A) and (R) are correct and (R) is the correct explanation of (A)
D
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Official Correct Answer:Option (B)
Step-by-Step Solution:
From left to right in periodic table :-
Metallic character decreases
Non-metallic character increases
⇒ It is due to increase in ionization enthalpy and increase in electron gain enthalpy.
Q 58ChemistrySection A • MCQgaseous state
+4 / -1
Which one of the following is the correct PV vs P plot at constant temperature for an ideal gas? (P and V stand for pressure and volume of the gas respectively)
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
PV = nRT (n, T constant)
PV = constant
Q 59ChemistrySection 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) : A simple distillation can be used to separate a mixture of propanol and propanone.
Reason (R) : Two liquids with a difference of more than 20∘ in their boiling points can be separated by simple distillations.
In the light of the above statements, choose the most appropriate answer from the options given below :
A
(A) is false but (R) is true.
B
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
C
(A) is true but (R) is false
D
Both (A) and (R) are correct and (R) is the correct explanation of (A).
Official Correct Answer:Option (D)
Step-by-Step Solution:
Simple distillation can be used to separate a
mixture of propanol and propanone.
• Two liquid with difference in B.P. around 20°C
can be separated by simple distillation.
• B.P. of propanol is 370 K.
B.P. of acetone is 329 K.
Both assertion & reason are correct & (R) is the correct explanation of (A).
Q 60ChemistrySection B • Numericalp block elements
+4 / -1
The number of halogen/(s) forming halic (V) acid is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 3
Official Correct Answer:3
Step-by-Step Solution:
Except F and At, all other halide can form Halic (V)
acid.
F cannot go in +5 oxidation state.
At is radioactive.
The number of halogen forming halic (V) acid
HClO3
HBrO3
HIO3
So answer is 3
Q 61MathematicsSection A • MCQstraight lines and pair of straight lines
+4 / -1
If p and q are the lengths of the perpendiculars from the origin on the lines,
x cosec α− y sec α = k cot 2α and
x sinα + y cosα = k sin2α
respectively, then k2 is equal to :
A
4p2 + q2
B
2p2 + q2
C
p2 + 2q2
D
p2 + 4q2
Official Correct Answer:Option (A)
Step-by-Step Solution:
First line is sinαx−cosαy=sin2αkcos2α
⇒xcosα−ysinα=2kcos2α
⇒p=2kcosα⇒2p=∣kcos2α∣ .... (i)
second line is xsinα+ycosα=ksin2α
⇒q=∣ksin2α∣ .... (ii)
Hence, 4p2+q2=k2 (From (i) & (ii))
Q 62MathematicsSection B • Numericalprobability
+4 / -1
An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 28
Official Correct Answer:28
Step-by-Step Solution:
I1 = first unit is functioning
I2 = second unit is functioning
P(I1) = 0.9, P(I2) = 0.8
P(I1) = 0.1, P(I2) = 0.2
P=0.1×0.2+0.9×0.2+0.1×0.80.8×0.1=288
98P=288×98=28
Q 63MathematicsSection B • Numericalapplication of derivatives
+4 / -1
If 'R' is the least value of 'a' such that the function f(x) = x2 + ax + 1 is increasing on [1, 2] and 'S' is the greatest value of 'a' such that the function f(x) = x2 + ax + 1 is decreasing on [1, 2], then the value of |R − S| is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
f(x) = x2 + ax + 1
f'(x) = 2x + a
when f(x) is increasing on [1, 2]
2x + a ≥ 0 ∀ x∈[1, 2]
a ≥−2x ∀ x∈[1, 2]
R = −4
when f(x) is decreasing on [1, 2]
2x + a ≤ 0 ∀ x∈[1, 2]
a ≤−2 ∀ x∈[1, 2]
S = −2
|R − S| = | −4 + 2 | = 2
Q 64MathematicsSection B • Numericalcomplex numbers
+4 / -1
A point z moves in the complex plane such that arg(z+2z−2)=4π, then the minimum value of z−92−2i2 is equal to _______________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 98
Official Correct Answer:98
Step-by-Step Solution:
Let z=x+iy
arg(x+2+iyx−2+iy)=4π
arg(x−2+iy)−arg(x+2+iy)=4π
tan−1(x−2y)−tan−1(x+2y)=4π
1+(x−2y).(x+2y)x−2y−x+2y=tan4π=1
x2−4+y2xy+2y−xy+2y=1
4y=x2−4+y2
x2+y2−4y−4=0
locus is a circle with center (0, 2) & radius = 22
min. value = (AP)2=(OP−OA)2
=(92−22)2
=(72)2=98
Q 65MathematicsSection A • MCQvector algebra
+4 / -1
Let a and b be two vectors such that 2a+3b=3a+b and the angle between a and b is 60∘. If 81a is a unit vector, then b is equal to :
Q 66MathematicsSection A • MCQdifferential equations
+4 / -1
If dxdy=2y2x+y−2x, y(0) = 1, then y(1) is equal to :
A
log2(2 + e)
B
log2(1 + e)
C
log2(2e)
D
log2(1 + e2)
Official Correct Answer:Option (B)
Step-by-Step Solution:
dxdy=2y2x+y−2x
2ydxdy=2x(2y−1)
∫2y−12ydy=∫2xdx
ln2ln(2y−1)=ln22x+C
⇒log2(2y−1)=2xlog2e+C
∵y(0)=1⇒0=log2e+C
C=−log2e
⇒log2(2y−1)=(2x−1)log2e
put x = 1, log2(2y−1)=log2e
2y = e + 1
y = log2(e + 1) Ans.
Q 67MathematicsSection B • Numericalpermutations and combinations
+4 / -1
The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 576
Official Correct Answer:576
Step-by-Step Solution:
Total possible words = 6! = 720
When 4 consonants are together (V, W, L, S)
such cases = 3! ⋅ 4! = 144
All consonants should not be together
= Total − All consonants together,
= 6! − 3! 4! = 576
Q 68MathematicsSection A • MCQheight and distance
+4 / -1
A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
A
1215
B
1210
C
810
D
610
Official Correct Answer:Option (B)
Step-by-Step Solution:
Let height of pole = 10l
tanα=183l=6l
tan2α=1810l
1−tan2α2tanα=1810l
use tanα=6l⇒l=572
height of pole = 10l=1210
Q 69MathematicsSection A • MCQsequences and series
Let the equation of the plane, that passes through the point (1, 4, −3) and contains the line of intersection of the planes 3x − 2y + 4z − 7 = 0 and x + 5y − 2z + 9 = 0, be αx + βy + γz + 3 = 0, then α + β + γ is equal to :
A
−23
B
−15
C
23
D
15
Official Correct Answer:Option (A)
Step-by-Step Solution:
3x − 2y + 4z − 7 + λ(x + 5y − 2z + 9) = 0
(3 + λ)x + (5λ− 2)y + (4 − 2λ)z + 9λ− 7 = 0
passing through (1, 4, −3)
⇒ 3 + λ + 20λ− 8 − 12 + 6λ + 9λ− 7 = 0
⇒λ = 32
⇒ equation of plane is
−11x − 4y − 8z + 3 = 0
⇒α + β + γ = −23
Q 71MathematicsSection A • MCQapplication of derivatives
+4 / -1
The number of real roots of the equation
e4x+2e3x−ex−6=0 is :
A
2
B
4
C
1
D
0
Official Correct Answer:Option (C)
Step-by-Step Solution:
Let ex=t>0
f(t)=t4+2t3−t−6=0
f′(t)=4t3+6t2−1
f′′(t)=12t2+12t>0
f(0)=−6,f(1)=−4,f(2)=24
⇒ Number of real roots = 1
Q 72MathematicsSection A • MCQlimits continuity and differentiability
The line 12xcosθ+5ysinθ=60 is tangent to which of the following curves?
A
x2 + y2 = 169
B
144x2 + 25y2 = 3600
C
25x2 + 12y2 = 3600
D
x2 + y2 = 60
Official Correct Answer:Option (B)
Step-by-Step Solution:
12xcosθ+5ysinθ=60
5xcosθ+12ysinθ=1
is tangent to 25x2+144y2=1
144x2+25y2=3600
Q 77MathematicsSection B • Numericaldefinite integration
+4 / -1
If xϕ(x)=5∫x(3t2−2ϕ′(t))dt, x > −2, and ϕ(0) = 4, then ϕ(2) is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
Official Correct Answer:4
Step-by-Step Solution:
xϕ(x)=5∫x3t2−2ϕ′(t)dt
xϕ(x)=x3−125−2[ϕ(x)−ϕ(5)]
xϕ(x)=x3−125−2ϕ(x)−2ϕ(5)
ϕ(0)=4⇒ϕ(5)=2133
ϕ(x)=x+2x3+8
ϕ(2)=4
Q 78MathematicsSection B • Numericalcircle
+4 / -1
If the variable line 3x + 4y = α lies between the two circles (x − 1)2 + (y − 1)2 = 1 and (x − 9)2 + (y − 1)2 = 4, without intercepting a chord on either circle, then the sum of all the integral values of α is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 165
Official Correct Answer:165
Step-by-Step Solution:
Both centers should lie on either side of the line as well as line can be tangent to circle.
(3 + 4 −α) . (27 + 4 −α) < 0
(7 −α) . (31 −α) < 0 ⇒α∈ (7, 31) ....... (1)
d1 = distance of (1, 1) from line
d2 = distance of (9, 1) from line
d1≥r1⇒5∣7−α∣≥1⇒α∈(−∞,2]∪[12,∞) .... (2)
d2≥r2⇒5∣31−α∣≥2⇒α∈(−∞,21]∪[41,∞) ....(3)
(1) ∩ (2) ∩ (3) ⇒α∈ [12, 21]
Sum of integers = 165
Q 79MathematicsSection A • MCQsets and relations
+4 / -1
Which of the following is not correct for relation R on the set of real numbers ?
A
(x, y) ∈ R ⇔ 0 < |x| − |y| ≤ 1 is neither transitive nor symmetric.
B
(x, y) ∈ R ⇔ 0 < |x − y| ≤ 1 is symmetric and transitive.
C
(x, y) ∈ R ⇔ |x| − |y| ≤ 1 is reflexive but not symmetric.
D
(x, y) ∈ R ⇔ |x − y| ≤ 1 is reflexive nd symmetric.
Official Correct Answer:Option (B)
Step-by-Step Solution:
Note that (a, b) and (b, c) satisfy 0 < |x − y| ≤ 1 but (a, c) does not satisfy it so 0 ≤ |x − y| ≤ 1 is symmetric but not transitive.
For example,
x = 0.2, y = 0.9, z = 1.5
0 ≤ |x – y| = 0.7 ≤ 1
0 ≤ |y – z| = 0.6 ≤ 1
But |x – z| = 1.3 > 1
So, (b) is correct.
Q 80MathematicsSection A • MCQmatrices and determinants
Q 81MathematicsSection B • Numericalbinomial theorem
+4 / -1
If (4436)k is the term, independent of x, in the binomial expansion of (4x−x212)12, then k is equal to ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 55
Official Correct Answer:55
Step-by-Step Solution:
(4x−x212)12
Tr+1=(−1)r.12Cr(4x)12−r(x212)r
Tr+1=(−1)r.12Cr(41)12−r(12)r.(x)12−3r
Term independent of x ⇒ 12 − 3r = 0 ⇒ r = 4
T5=(−1)r.12Cr(41)8(12)4=4436.k
⇒ k = 55
Q 82MathematicsSection A • MCQindefinite integration
+4 / -1
The integral ∫\root4\of(x−1)3(x+2)51dx is equal to : (where C is a constant of integration)
A
43(x−1x+2)41+C
B
43(x−1x+2)45+C
C
34(x+2x−1)41+C
D
34(x+2x−1)45+C
Official Correct Answer:Option (C)
Step-by-Step Solution:
∫(x−1)3/4(x+2)5/4dx
=∫(x−1x+2)5/4.(x−1)2dx
put x−1x+2=t
=−31∫t5/4dt
=34.t1/41+C
=34(x+2x−1)1/4+C
Q 83MathematicsSection A • MCQmathematical reasoning
+4 / -1
Let *, ▢ ∈{∧, ∨} be such that the Boolean expression (p * ∼ q) ⇒ (p ▢ q) is a tautology. Then :
A
* = ∨, ▢ = ∨
B
* = ∧, ▢ = ∧
C
* = ∧, ▢ = ∨
D
* = ∨, ▢ = ∧
Official Correct Answer:Option (C)
Step-by-Step Solution:
(p ∧∼ q) → (p ∨ q) is tautology
p
q
∼ q
p ∧∼ q
p ∨ q
(p ∧∼ q) → (p ∨ q)
T
T
F
F
T
T
T
F
T
T
T
T
F
T
F
F
T
T
F
F
T
F
F
T
Q 84MathematicsSection A • MCQsequences and series
+4 / -1
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r2, then r2− d is equal to :
A
7 − 73
B
7 + 3
C
7 −3
D
7 + 33
Official Correct Answer:Option (B)
Step-by-Step Solution:
Let numbers be ra, a, ar → G.P.
ra, 2a, ar → A.P. ⇒ 4a = ra + ar ⇒ r + r1 = 4
r = 2 ±3
4th form of G.P. = 3r2⇒ ar2 = 3r2⇒ a = 3
r = 2 + 3, a = 3, d = 2a −ra = 33
r2− d = (2 + 3)2− 33
= 7 + 43− 33
= 7 + 3
Q 85MathematicsSection A • MCQmatrices and determinants
+4 / -1
If ar=cos92rπ+isin92rπ, r = 1, 2, 3, ....., i = −1, then the determinant \left| {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right| is equal to :
A
a2a6− a4a8
B
a9
C
a1a9− a3a7
D
a5
Official Correct Answer:Option (C)
Step-by-Step Solution:
ar=e9i2πr, r = 1, 2, 3, ......, a1, a2, a3, ..... are in G.P.
Q 87MathematicsSection A • MCQdefinite integration
+4 / -1
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫0x1−(f′(t))2dt=∫0xf(t)dt, 0≤x≤1 and f(0) = 0, then x→0limx21∫0xf(t)dt :
A
equals 0
B
equals 1
C
does not exist
D
equals 21
Official Correct Answer:Option (D)
Step-by-Step Solution:
∫0x1−(f′(t))2dt=∫0xf(t)dt,0≤x≤1
differentiating both the sides
1−(f′(x))2=f(x)
⇒1−(f′(x))2=f2(x)
1−f2(x)f′(x)=1
sin−1f(x)=x+C
∵f(0)=0⇒C=0⇒f(x)=sinx
Now, x→0limx20∫xsintdt(00)=21
Q 88MathematicsSection B • Numericalstatistics
+4 / -1
The mean of 10 numbers 7 × 8, 10 × 10, 13 × 12, 16 × 14, ....... is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 398
Official Correct Answer:398
Step-by-Step Solution:
7 × 8, 10 × 10, 13 × 12, 16 × 14 ........
Tn = (3n + 4) (2n + 6) = 2(3n + 4) (n + 3)
= 2(3n2 + 13n + 12) = 6n2 + 26n + 24
S10 = n=1∑10Tn=6n=1∑10n2+26n=1∑10n+24n=1∑101
=66(10×11×21)+26×210×11+24×10
=10×11(21+13)+240
= 3980
Mean =10S10=103980=398
Q 89MathematicsSection B • Numericaldefinite integration
+4 / -1
Let [t] denote the greatest integer ≤ t. Then the value of
8.−21∫1([2x]+∣x∣)dx is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
I=−21∫1([2x]+∣x∣)dx
=−1/2∫1[2x]dx+−1/2∫1∣x∣dx
=0+−1/2∫0(−x)dx+0∫1xdx
=(−2x2)−1/20+(2x2)01
=(0+81)+21
=85
∴ 8I = 5
Q 90MathematicsSection A • MCQparabola
+4 / -1
The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (> R) respectively from the origin, is :
