Official NTA Question PaperFull Solved Paper2021Morning Shift
JEE Main 2021 25th February 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.
An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :
A
2u+v
B
2v2−u2
C
2v−u
D
2v2+u2
Official Correct Answer:Option (D)
Step-by-Step Solution:
Let initial speed of train u. When midpoint of the train reach the signal post it's velocity becomes v0.
∴v02=u2+2as .......(1)
When train passes the signal post completely it's velocity becomes v.
∴v2=v02+2as ......(2)
Subtracting (2) from (1) we get,
v02−v2=u2−v02
⇒v02+v02=u2+v2
⇒v0=2u2+v2
Q 2PhysicsSection B • Numericalcircular motion
+4 / -1
A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s2)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5
Official Correct Answer:5
Step-by-Step Solution:
Let the speed of bob at lowest position be v1 and at the highest position be v2.
Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,
21mv12=21mv22+mg2l
⇒v12=v22+4gl ..........(1)
Now, Tmax−mg=lmv12
⇒Tmax=mg+lmv12
& Tmin+mg=lmv22
⇒Tmin=lmv22−mg
TminTmax=15
⇒lmv22−mgmg+lmv12=15
⇒mg+lmv12=[lmv22−mg]5
⇒mg+lm[v22+4gl]=l5mv22−5mg
⇒mg+lmv22+4mg=l5mv22−5mg
⇒10mg=l4mv22
v22=410×10×1
⇒v22=25⇒v2=5 m/s
Thus, velocity of bob at highest position 5 m/s.
Q 3PhysicsSection A • MCQunits and measurements
+4 / -1
Match List - I with List - II :
List I
List II
(a)
h (Planck's constant)
(i)
[MLT−1]
(b)
E (kinetic energy)
(ii)
[ML2T−1]
(c)
V (electric potential)
(iii)
[ML2T−2]
(d)
P (linear momentum)
(iv)
[ML2I−1T−3]
Choose the correct answer from the options given below :
A
(a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)
B
(a) → (i), (b) → (ii), (c) → (iv), (d) → (iii)
C
(a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)
D
(a) → (iii), (b) → (iv), (c) → (ii), (d) → (i)
Official Correct Answer:Option (A)
Step-by-Step Solution:
Kinetic Energy,
21mv2=[ML2T−2]
Momentum,
mv=[MLT−1]
Plank constant :
E=hγ
⇒ML2T−2=h×T1
⇒h=[ML2T−1]
Also, E=qV
⇒V=[C][ML2T−2]=[ML2T−2C−1]
Q 4PhysicsSection A • MCQsimple harmonic motion
+4 / -1
If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
A
16 m/s2
B
2π2 ms−2
C
π2 ms−2
D
9.8 ms−2
Official Correct Answer:Option (B)
Step-by-Step Solution:
The formula for the period of a simple pendulum is given by:
T=2πgl
where:
T is the period of the pendulum,
l is the length of the pendulum, and
g is the acceleration due to gravity.
We need to find g. Rearranging the formula for g, we get:
g=T24π2l
Given:
l = 2 m,
T = 2 s,
Substituting these values into the equation, we get:
g=(2)24π2×2=2π2ms−2
Q 5PhysicsSection A • MCQheat and thermodynamics
+4 / -1
A diatomic gas, having Cp=27R and Cv=25R, is heated at constant pressure. The ratio dU : dQ : dW :
A
5 : 7 : 3
B
3 : 7 : 2
C
5 : 7 : 2
D
3 : 5 : 2
Official Correct Answer:Option (C)
Step-by-Step Solution:
dV=n25RΔT
dQ=n27RΔT
dW=dQ−dV
=n22RΔT
∴dV:dQ:dW
=n25RΔT:n27RΔT:n22RΔT
=5:7:2
Q 6PhysicsSection B • Numericalray optics
+4 / -1
The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is __________ cm.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 15
Official Correct Answer:15
Step-by-Step Solution:
v1−u1=f1 .... (1)
m = uv ..... (2)
from (1) and (2) we get
m = f+uf
given conditions
m1 = -m2
f−10f=f−20−f
⇒ f – 20 = -f + 10
⇒ 2f = 30
⇒ f = 15 cm
Q 7PhysicsSection A • MCQalternating current
+4 / -1
The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
A
0.8 H and 150 μF
B
0.8 H and 250 μF
C
1.33 H and 150 μF
D
1.33 H and 250 μF
Official Correct Answer:Option (B)
Step-by-Step Solution:
Since key is open, circuit is series
15 = irms (60)
∴ irms = 41A
Now, 20 = 41 XL = 41 (ωL)
∴ L = 54 = 0.8 H
and 10 = 41(100C)1
C = 40001F = 250μF
Q 8PhysicsSection B • Numericalcurrent electricity
+4 / -1
In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of E2E1 is ba, where a = _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
Length of AB = 10 m
For battery E1, balancing length is l1
l1 = 380 cm [from end A]
For battery E2, balancing length is l2
l2 = 760 cm [from end A]
Now, we know that E2E1=l2l1
⇒E2E1=760380=21=ba
∴ a = 1 & b = 2
∴ a = 1
Q 9PhysicsSection A • MCQmagnetics
+4 / -1
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ________.
A
1.0 m
B
0.15 m
C
0.2 m
D
0.1 m
Official Correct Answer:Option (D)
Step-by-Step Solution:
B=2(R2+x2)3/2μ0NiR2
at x1 = 0.05 m, B1=2(R2+(0.05)2)3/2μ0NiR2
at x2 = 0.2 m, B2=2(R2+(0.2)2)3/2μ0NiR2
B2B1=(R2+0.0025)3/2(R2+0.04)3/2
(18)2/3=R2+0.0025R2+0.04
4 (R2 + 0.0025) = R2 + 0.04
3R2 = 0.04 − 0.0100
R2 = 30.03 = 0.01
R = 0.01 = 0.1 m
Q 10PhysicsSection A • MCQvector algebra
+4 / -1
In an octagon ABCDEFGH of equal side, what is the sum of
AB+AC+AD+AE+AF+AG+AH,
if, AO=2i+3j−4k
A
−16i−24j+32k
B
16i+24j−32k
C
16i+24j+32k
D
16i−24j+32k
Official Correct Answer:Option (B)
Step-by-Step Solution:
We know,
∵OA+OB+OC+OD+OE+OF+OG+OH=0
By triangle law of vector addition, we can write
AB=AO+OB;AC=AO+OC
AD=AO+OD;AE=AO+OE
AF=AO+OF;AG=AO+OG
AH=AO+OH
Now
AB+AC+AD+AE+AF+AG+AH
=(7AO)+OB+OC+OD+OE+OF+OG+OH
=(7AO)+0−OA
=(7AO)+AO
=8AO=8(2i+3j−4k)
=16i+24j−32k
Q 11PhysicsSection B • Numericalelectrostatics
+4 / -1
The electric field in a region is given by E=(53E0i+54E0j)CN. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y − z plane) to that of the surface of area 0.3 m2 (parallel to x − z plane) is a : b, where a = __________ [Here i, j and k are unit vectors along x, y and z-axes respectively.]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
ϕ=E.A
Aa=0.2i
Ab=0.3j
ϕa=(53E0i+54E0j).0.2i
⇒ϕa=53E0×0.2
ϕb=(53E0i+54E0j).0.3j
⇒ϕb=54E0×0.3
ba=ϕbϕa=54E0×0.353E0×0.2=126=0.5
Q 12PhysicsSection B • Numericalwork power and energy
+4 / -1
The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as
U=r10α−r5β−3
where, α and β are positive constants. The equilibrium distance between two atoms will be (β2α)ba, where a = ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
F=−drdU
F=−[−r1110α+r65β]
for equilibrium, F = 0
r1110α=r65β
β2α=r5
r=(β2α)1/5
∴a=1
Q 13PhysicsSection A • MCQcommunication systems
+4 / -1
Given below are two statements :
Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.
Statement II : The side band frequencies are 1002 kHz and 998 kHz.
In the light of the above statements, choose the correct answer from the options given below :
A
Statement I is false but Statement II is true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Both Statement I and Statement II are true
Official Correct Answer:Option (D)
Step-by-Step Solution:
Side band = (fc− fm) to (fc + fm)
= (1000 − 2) KHz to (1000 + 2) Khz
= 998 KHz to 1002 KHz
Band width = 2fm
= 2 × 2 KHz
= 4 KHz
Both statements are true.
Q 14PhysicsSection A • MCQalternating current
+4 / -1
The current (i) at time t = 0 and t = ∞ respectively for the given circuit is :
A
5518E,185E
B
185E,5518E
C
185E,3310E
D
3310E,185E
Official Correct Answer:Option (C)
Step-by-Step Solution:
at t = 0, inductor is removed, so circuit will look like this
at t = 0
Req=6+96×9=1554
I(t=0)=54E×15=185E
at t = ∞, inductor is replaced by plane wire, so circuit will look like this
at t = ∞,
I(t=∞)=25+54E=3310E
Now,
Req=1+41×4+5+55×5
=54+25=108+25=1033
I=ReqE=3310E
Q 15PhysicsSection A • MCQelectronic devices
+4 / -1
A 5V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.
A
∼ 0.5 A
B
∼ 1.5 A
C
∼ 0.43 A
D
∼ 0.86 A
Official Correct Answer:Option (C)
Step-by-Step Solution:
Since silicon diode is used so 0.7 Volt is drop across it, only D1 will conduct.
So current through cell,
I = 105−0.7 = 0.43 A
Q 16PhysicsSection A • MCQgravitation
+4 / -1
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass.
Reason R : The product of their mass and radius must be same. M1R1 = M2R2
In the light of the above statements, choose the most appropriate answer from the options given below :
A
Both A and R are correct and R is the correct explanation of A
B
Both A and R are correct but R is NOT the correct explanation of A
C
A is correct but R is not correct
D
A is not correct but R is correct
Official Correct Answer:Option (C)
Step-by-Step Solution:
ve = escape velocity
ve=R2GM
so for same ve,R1M1=R2M2
A is true but R is false
Q 17PhysicsSection A • MCQatoms and nuclei
+4 / -1
Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio N2N1 will be equal to :
A
81
B
13
C
31
D
18
Official Correct Answer:Option (D)
Step-by-Step Solution:
Let Half life of x = t
then half life of y = 2t
when 3 half life of y is completed then 6 half life of x is completed.
∴ Now x have = 26N1 nuclei
and y have = 23N2 nuclei
From question,
26N1=23N2
⇒N2N1=8
Q 18PhysicsSection B • Numericalelectromagnetic induction
+4 / -1
A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______J.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 144
Official Correct Answer:144
Step-by-Step Solution:
Ldtdi=ε
=3t
L∫di=3∫tdt
Li=23t2
i=2L3t2
energy, E=21Li2
=21L(2L3t2)2
=21×4L9t4
=89×4×2(4)4=144 J
Q 19PhysicsSection B • Numericalheat and thermodynamics
+4 / -1
In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 100∘C to 300∘C will be ___________ nR, where n denotes number of moles of a gas.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 50
Official Correct Answer:50
Step-by-Step Solution:
P=kv3
⇒pv−3=k
⇒x=−3
w=x−1nR(T1−T2)
=−3−1nR(100−300)
=−4nR(−200)
=50 nR
Q 20PhysicsSection A • MCQwave optics
+4 / -1
Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio Imax+IminImax−Imin will be :
A
2x+122x
B
x+122x
C
x+12x
D
2x+12x
Official Correct Answer:Option (A)
Step-by-Step Solution:
Given, I2I1=2x
We know,
IminImax=(I1−I2I1+I2)2
=(I2I1−1I2I1+1)2
=(2x−12x+1)2
Now,
Imax+IminImax−Imin
=IminImax+1IminImax−1
=(2x−12x+1)2+1(2x−12x+1)2−1
=(2x+1)2+(2x−1)2(2x+1)2−(2x−1)2
=2+4x42x=1+2x22x
Q 21PhysicsSection A • MCQunits and measurements
+4 / -1
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :
A
1.80 mm
B
0.90 mm
C
0.82 mm
D
1.64 mm
Official Correct Answer:Option (C)
Step-by-Step Solution:
Least count = 1001mm=0.01mm
zero error = + 8 × LC = + 0.08 mm
True reading (Diameter)
= (1 mm + 72 × LC) − (Zero error)
= (1 mm + 72 × 0.01 mm) − 0.08 mm
= 1.72 mm − 0.08 mm
= 1.64 mm
Therefore, radius = 21.64 = 0.82 mm.
Q 22PhysicsSection A • MCQmagnetics
+4 / -1
A proton, a deuteron and an α particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is _________ and their speed is _______, in the ratio.
A
2 : 1 : 1 and 4 : 2 : 1
B
4 : 2 : 1 and 2 : 1 : 1
C
1 : 2 : 4 and 2 : 1 : 1
D
1 : 2 : 4 and 1 : 1 : 2
Official Correct Answer:Option (A)
Step-by-Step Solution:
F = qvB = qmpB
F ∝mq [as p, B are const.]
∴ F1 : F2 : F3
= me:2me:4m2e
= 2 : 1 : 1
And v1 : v2 : v3
= mp:2mp:4mp
= 4 : 2 : 1
Q 23PhysicsSection B • Numericalheat and thermodynamics
+4 / -1
A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is 3Rx. Value of x is ___________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 3600
Official Correct Answer:3600
Step-by-Step Solution:
ΔKE=ΔU
ΔU=nCvΔT
21mv2=23nRΔT
3nRmv2=ΔT
3×1×R4×(30)2=ΔT
⇒ΔT=R1200
⇒3Rx=R1200
⇒x=3600
Q 24PhysicsSection A • MCQgravitation
+4 / -1
Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.
If TA and TB are the time periods of A and B respectively then the value of TB− TA :
[Given : radius of earth = 6400 km, mass of earth = 6 × 1024 kg]
A
1.33 × 103 s
B
4.24 × 102 s
C
3.33 × 102 s
D
4.24 × 103 s
Official Correct Answer:Option (A)
Step-by-Step Solution:
T=2πGMr3
TA=2πGM(6400+600)3
TA=2π×109GM73
TB=2π×109GM83
TB−TA=GM2π109[88−77]
=314×4.107
=1289.64
=1.289×103s
Q 25PhysicsSection A • MCQgravitation
+4 / -1
A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius (2R) is made in the sphere (as shown in figure) and the force becomes F2. The value of F1 : F2 is
A
36 : 25
B
41 : 50
C
50 : 41
D
25 : 36
Official Correct Answer:Option (C)
Step-by-Step Solution:
g1=(3R)2GM=9R2GM
g2=9R2GM−(3R−2R)2G(8M)
=9R2GM−R250GM=9×5041R2GM
g2g1=5041
∴F2F1=mg2mg1=5041
Q 26PhysicsSection A • MCQwaves
+4 / -1
A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :
A
13 cm
B
18.4 cm
C
16.6 cm
D
14.8 cm
Official Correct Answer:Option (D)
Step-by-Step Solution:
∴l+1.8=4λ
Also λ=fv=504336
⇒l+1.8=4×504336
⇒l=14.86 cm
Q 27PhysicsSection A • MCQdual nature of radiation
+4 / -1
An α particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λαλp is :
A
8
B
2.8
C
7.8
D
3.8
Official Correct Answer:Option (B)
Step-by-Step Solution:
We know,
qv=2mp2
⇒p=2mqv
∴λ=2mqvh
∴λαλp=2mpqpv2mαqαv
=2me2(4m)(2e)
=8
=22
Q 28PhysicsSection B • Numericalelectrostatics
+4 / -1
512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 128
Official Correct Answer:128
Step-by-Step Solution:
Let charge on each drop = q
radius = r
v=rkq
⇒2=rkq
radius of bigger
34πR3=512×34πr3
R=8r
∴v=Rk(512)q=8512rkq=8512×2
=128V
Q 29PhysicsSection B • Numericalalternating current
+4 / -1
A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 μF is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ × 10−8 H.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 10
Official Correct Answer:10
Step-by-Step Solution:
λ = 960 m
C = 2.56 μF = 2.56 × 10−6 F
c = 3 × 108 m/s
L = ?
Now at resonance, ω0=LC1
[Resonant frequency]
2πf0=LC1
On substituting f0=λc, we have 2πλc=LC1
Squaring both sides : 4π2λ2c2=LC1
=(960)24×10×(3×108)2=L×2.56×10−61
⇒L1=960×9604×10×9×1016×2.56×10−6
⇒L=10×10−8 H
Q 30PhysicsSection A • MCQheat and thermodynamics
+4 / -1
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.
Reason R : On heating, the length of the rod increases.
In the light of the above statements, choose the correct 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 B are true but R is NOT the correct explanation of A
D
Both A and R are true and R is the correct explanation of A
Official Correct Answer:Option (C)
Step-by-Step Solution:
When a rod is free and it is heated then there is no thermal stress produced in it.
The rod will expand due to increase in temperature.
So both A & R are true.
Q 31ChemistrySection B • Numericalredox reactions
+4 / -1
0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with 10N HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is ________. (Rounded 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:
1st end point reaction :
⟶NaCl+H2ONaOH+HCl\buildrel
nf=1
⟶NaHCO3NaCO3+HCl\buildrel
nf=1
Eq of HCl used =nNaOH×1+nNa2CO3×1
17.5×101×10−3=nNaOH+nNa2CO3
2nd end point :
⟶H2CO3NaHCO3+HCl\buildrel
1.5×101×10−3=nNaHCO3×1=nNaHCO3
0.15 mmol = nNa2CO3
0.15=nNa2CO3
wNa2CO3=0.40.15×106×10−3×100×10
= 3.975%
≃ 4%
Q 32ChemistrySection A • MCQsome basic concepts of chemistry
+4 / -1
Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :
A
51.63
B
50.33
C
63.53
D
53.33
Official Correct Answer:Option (D)
Step-by-Step Solution:
CxHyOz + O2→ xCO2 + 2yH2O
2.64 g of CO2 contains 0.72 g C.
1.08 g of H2O contains 0.12 g H.
∴ Mass of oxygen present = 1.80 – (0.72
+0.12) = 0.96 g
% of O = 1.800.96×100 = 53.33 %
Q 33ChemistrySection A • MCQhaloalkanes and haloarenes
+4 / -1
Identify A and B in the chemical reaction.
A
B
C
D
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 34ChemistrySection B • Numericalchemical kinetics and nuclear chemistry
+4 / -1
For the reaction, aA + bB → cC + dD, the plot of log k vs T1 is given below :
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 526
Official Correct Answer:526
Step-by-Step Solution:
log10K=log10A−2.303RTEa
Slope =2.303REa=−10000
log10K1K2=2.303REa×[T11−T21]
log1010−510−4=10000×[5001−T1]
⇒1=10000×[5001−T1]
⇒100001=5001−T1
⇒T1=5001−100001
⇒T1=1000020−1=1000019
⇒T=1910000=526 K
Q 35ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
The major product of the following chemical reaction is :
A
CH3CH2CHO
B
CH3CH2CH3
C
CH3CH2CH2OH
D
(CH3CH2CO)2O
Official Correct Answer:Option (A)
Step-by-Step Solution:
Q 36ChemistrySection B • Numericalthermodynamics
+4 / -1
The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol−1, while the electron gain enthalpy of Br is −325.0 kJ mol−1. Given the lattice enthalpy of NaBr is −728.4 kJ mol−1. The energy for the formation of NaBr ionic solid is (−) ____________ × 10−1 kJ mol−1.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 5576
Official Correct Answer:5576
Step-by-Step Solution:
⟶Na+(g)Na(s)\buildrel
ΔH=495.8
⟶Br−(g)21Br2(l)+e−\buildrel
ΔH=325
⟶NaBr(s)Na+(g)+Br−(g)\buildrel
ΔH=−728.4
⟶NaBr(s).Na(s)+21Br2(l)\buildrel
ΔH=?
ΔH=495.8−325−728.4−557.6 kJ
=−5576×10−1 kJ
Q 37ChemistrySection A • MCQionic equilibrium
+4 / -1
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 × 10−16 and Ka(HCN) = 6.2 × 10−10]
A
1.9 × 10−5
B
1.6 × 10−6
C
2.2 × 10−16
D
0.625 × 10−6
Official Correct Answer:Option (A)
Step-by-Step Solution:
Let solubility is x
Ksp×Ka1=[Ag+][CN−]×[H+][CN−][HCN]
2.2×10−16×6.2×10−101=10−3[S][S]
S2=6.22.2×10−9
S2=3.55×10−10
S=3.55×10−10
S=1.88×10−5=1.9×10−5
Q 38ChemistrySection A • MCQp block elements
+4 / -1
The correct statement about B2H6 is :
A
Its fragment, BH3, behaves as a Lewis base.
B
All B−H−B angles are of 120∘.
C
The two B−H−B bonds are not of same length.
D
Terminal B−H bonds have less p-character when compared to bridging bonds.
Official Correct Answer:Option (D)
Step-by-Step Solution:
Terminal B – H bonds are shorter than the
bridging B – H bonds which shows that the
terminal B – H bonds have greater s-character
and less p-character.
Q 39ChemistrySection A • MCQhydrocarbons
+4 / -1
Identify A in the given chemical reaction.
A
B
C
D
Official Correct Answer:Option (D)
Step-by-Step Solution:
Mo2O3
at 773 K temperature and 10-20-atm
pressure is aromatising agent.
Q 40ChemistrySection B • Numericalhydrocarbons
+4 / -1
Consider the following chemical reaction.
The number of sp2 hybridized carbon atom(s) present in the product is ________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 7
Official Correct Answer:7
Step-by-Step Solution:
All the 7-carbon-atoms in product are sp2
hybridised.
Q 41ChemistrySection A • MCQd and f block elements
+4 / -1
In which of the following pairs, the outer most electronic configuration will be the same?
A
Cr+ and Mn2+
B
Ni2+ and Cu+
C
V2+ and Cr+
D
Fe2+ and Co+
Official Correct Answer:Option (A)
Step-by-Step Solution:
Cr+→ [Ar]3d5
Mn2+→ [Ar]3d5
Q 42ChemistrySection A • MCQchemical bonding and molecular structure
+4 / -1
According to molecular orbital theory, the species among the following that does not exist is :
A
O22−
B
Be2
C
He2−
D
He2+
Official Correct Answer:Option (B)
Step-by-Step Solution:
Note :
According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
We know, Bond order =21 [Nb − Na]
Nb = No of electrons in bonding molecular orbital
Na= No of electrons in anti bonding molecular orbital
(4) upto 14 electrons, molecular orbital configuration is
Here Na = Anti bonding electron = 4 and Nb = 10
(5) After 14 electrons to 20 electrons molecular orbital configuration is - - -
Here Na = 10
and Nb = 10
Molecular orbital configuration of O 22− (18 electrons) is
Configuration of He2+ (3 electrons) is = σ1s2σ1s1∗
∴ Bond order = 21 (2 −1) = 0.5
Configuration of He2− (5 electrons) is = σ1s2σ1s2∗σ2s1
∴ Bond order = 21 (3 −2) = 0.5
Configuration of Be2 (4 electrons) is = σ1s2σ1s2∗
∴ Bond order = 21 (2 − 2) = 0
Q 43ChemistrySection B • Numericalsolutions
+4 / -1
1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm is _________ K. (Rounded off to the nearest integer)
[Given Kb for (H2O) = 0.52 K kg mol−1]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 375
Official Correct Answer:375
Step-by-Step Solution:
⟶2A+3+3B−2A2B3\buildrel
No. of ions = 2 + 3 = 5
i = 1 + (n − 1) α
= 1 + (5 − 1) × 0.6
= 1 + 4 × 0.6 = 1 + 2.4 = 3.4
ΔTb=Kb×m×i=0.52×1×3.4=1.768∘ C
ΔTb=(Tb)solution−[(Tb)H2O]Solution
1.768=(Tb)solution−100
(Tb)solution=101.768∘ C
=375 K
Q 44ChemistrySection B • Numericalthermodynamics
+4 / -1
The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be −742.24 kJ mol−1. The magnitude of ΔH298 for the reaction
NH2CH(S)+23O2(g)→N2(g)+O2(g)+H2O(I)
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol−1 K−1]
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 741
Which of the glycosidic linkage between galactose and glucose is present in lactose?
A
C-1 of galactose and C-6 of glucose
B
C-1 of glucose and C-6 of galactose
C
C-1 of galactose and C-4 of glucose
D
C-1 of glucose and C-4 of galactose
Official Correct Answer:Option (C)
Step-by-Step Solution:
Glycosidic linkage is a type of covalent bond that joins
carbohydrate molecules to another group.
In lactose, the glycosidic linkage is formed between C-1 of galactose
and C-4, glycosidic of glucose.
Q 46ChemistrySection A • MCQenvironmental chemistry
+4 / -1
Given below are two statements :
Statement I : An allotrope of oxygen is an important intermediate in the formation of reducing smog.
Statement II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog.
In the light of the above statements, choose the correct answer from the options given below :
A
Statement I is false but Statement II is true
B
Both Statement I and Statement II are false
C
Both Statement I and Statement II are true
D
Statement I is true but Statement II is false
Official Correct Answer:Option (B)
Step-by-Step Solution:
Reducing smog as is acts as reducing agent, the reducing character is due to presence of
sulphur dioxide and carbon particles.
Q 47ChemistrySection A • MCQhydrogen
+4 / -1
Which of the following equation depicts the oxidizing nature of H2O2?
A
Cl2+H2O2→2HCl+O2
B
KIO4+H2O2→KIO3+H2O+O2
C
2I−+H2O2+2H+→I2+2H2O
D
I2+H2O2+2OH−→2I−+2H2O+O2
Official Correct Answer:Option (C)
Step-by-Step Solution:
2I−+H2O2+2H+→I2+2H2O
Oxygen reduces from –1 to –2.
So, its reduction will takes place. Hence it will behave as oxidising agent or it shows oxidising
nature.
While in other option it change from (–1) to 0.
Q 48ChemistrySection A • MCQatomic structure
+4 / -1
The plots of radial distribution functions for various orbitals of hydrogen atom against 'r' are given below :
The correct plot for 3s orbital is :
A
(B)
B
(A)
C
(C)
D
(D)
Official Correct Answer:Option (D)
Step-by-Step Solution:
Number of radial nodes = n – (l + 1)
For 3s orbital n = 3, l = 0
Number of radial nodes = 3 – 0 – 1 = 2
It is correctly represented in graph of option D.
Q 49ChemistrySection B • Numericalsome basic concepts of chemistry
+4 / -1
In basic medium CrO42− oxidises S2O32− to form SO42− and itself changes into Cr(OH)4−. The volume of 0.154 M CrO42− required to react with 40 mL of 0.25 M S2O32− is __________ mL. (Rounded off to the nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 173
Q 50ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
Which one of the following reactions will not form acetaldehyde?
A
CH3CH2OH573K⟶Cu
B
⟶CH3CH2OH\buildrelCrO3−H2SO4
C
CH2=CH2+O2H2O⟶Pd(II)/Cu(II)
D
CH3CNii)H2O⟶i)DIBAL−H
Official Correct Answer:Option (B)
Step-by-Step Solution:
Q 51ChemistrySection A • MCQd and f block elements
+4 / -1
Given below are two statements :
Statement I : CeO2 can be used for oxidation of aldehydes and ketones.
Statement II : Aqueous solution of EuSO4 is a strong reducing agent.
In the light of the above statements, choose the correct 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 false
D
Both Statement I and Statement II are true
Official Correct Answer:Option (D)
Step-by-Step Solution:
The +3 oxidation state of lanthanide is most
stable and therefore lanthanide in +4 oxidation
state has strong tendence to gain e–
and
converted into +3 and therefore act as strong
oxidizing agent.
eg Ce+4 And therefore CeO2
is used to oxidized alcohol
aldehyde and ketones.
Lanthanide in +2 oxidation state has strong
tendency to loss e–
and converted into +3
oxidation state therefore act as strong reducing
agent.
∴ EuSO4 act as a strong reducing agent.
Q 52ChemistrySection A • MCQcoordination compounds
+4 / -1
The hybridization and magnetic nature of [Mn(CN)6]4− and [Fe(CN)6]3−, respectively are :
A
sp3d2 and diamagnetic
B
d2sp3 and paramagnetic
C
sp3d2 and paramagnetic
D
d2sp3 and diamagnetic
Official Correct Answer:Option (B)
Step-by-Step Solution:
[Mn(CN)6]4−
Mn2+ = 3d5 4s0
CN– is a strong field ligand.
∴ Pairing will occur.
[Fe(CN)6]3−
Fe3+ = 3d54s0
CN– is a strong field ligand.
∴ Pairing will occur.
Q 53ChemistrySection A • MCQaldehydes ketones and carboxylic acids
+4 / -1
Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are :
A
C only
B
B only
C
B and C only
D
A and B only
Official Correct Answer:Option (C)
Step-by-Step Solution:
The reactions of given compound with sodium bicarbonate
solution are as follows
As H2CO3 is weak acid it dissociate to liberate CO2 gas and H2O.
Equilibrium favours forward direction and CO2 is liberated. In the above two reactions, H2CO3 is comparatively weak acid.
∴CO2 gas is liberated.
Equilibrium favours backward direction and CO2 is not liberated.
Thus, only B and C will liberate carbon dioxide with sodium bicarbonate solution.
Q 54ChemistrySection A • MCQcompounds containing nitrogen
+4 / -1
Which of the following reaction/s will not give p-aminoazobenzene?
A
C only
B
A only
C
A and B
D
B only
Official Correct Answer:Option (D)
Step-by-Step Solution:
Q 55ChemistrySection A • MCQsurface chemistry
+4 / -1
In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption (mx) is directly proportional to Px. The value of x is :
A
∞
B
zero
C
1
D
n1
Official Correct Answer:Option (D)
Step-by-Step Solution:
According to Freundlich adsorption isotherm, in the median range of pressure
mx∝Pn1
⇒mx = kPn1
Hence, x = n1
Q 56ChemistrySection B • Numericalgaseous state
+4 / -1
A car tyre is filled with nitrogen gas at 35 psi at 27∘C. It will burst if pressure exceeds 40 psi. The temperature in ∘C at which the car tyre will burst is _________. (Rounded off to the nearest integer)
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 70
Official Correct Answer:70
Step-by-Step Solution:
T1P1=T2P2
30035=T240
T2=3540×300
=342.86 K
=69.85∘ C ≃70∘ C
Q 57ChemistrySection B • Numericalchemical bonding and molecular structure
+4 / -1
Among the following, the number of halide(s) which is/are inert to hydrolysis is _________.
(A) BF3
(B) SiCl4
(C) PCl5
(D) SF6
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 1
Official Correct Answer:1
Step-by-Step Solution:
BF3 – Shows Partial hydrolysis
SiCl4 – Undergoes hydrolysis readily
PCl5 – Undergoes hydrolysis by addition–
elimination mechanism.
SF6 – Due to crowding Inert towards hydrolysis.
Q 58ChemistrySection A • MCQisolation of elements
+4 / -1
Ellingham diagram is a graphical representation of :
A
ΔG vs P
B
ΔG vs T
C
(ΔG − TΔS) vs T
D
ΔH vs T
Official Correct Answer:Option (B)
Step-by-Step Solution:
Ellingham diagram is a graphical
representation of ΔG vs T when metal heated
with oxygen to form metal oxide.
Q 59ChemistrySection B • Numericalbasics of organic chemistry
+4 / -1
Using the provided information in the following paper chromatogram :
Fig : Paper chromatography for compounds A and B.
the calculated Rf value of A ________× 10-1.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 4
On chromatogram distance travelled by compound is = 2 cm
Distance travelled by solvent = 5 cm
So, Rf=52= 4 × 10−1 = 0.4
Q 60ChemistrySection A • MCQpolymers
+4 / -1
Which statement is correct?
A
Buna-N is a natural polymer.
B
Synthesis of Buna-S needs nascent oxygen.
C
Buna-S is a synthetic and linear thermosetting polymer.
D
Neoprene is an addition copolymer used in plastic bucket manufacturing.
Official Correct Answer:Option (B)
Step-by-Step Solution:
Buna-S is an elastomer
Buna-N is a synthetic polymer
Buna-S is polymerised by addition
polymerisation method which needs
radical initiator for chain propagation step.
Nascent oxygen can be used as an Radical
initiator.
Neoprene is a synthetic rubber.
Q 61MathematicsSection B • Numericalapplication of derivatives
+4 / -1
Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = −1 and x = 1. If x→0limx3f(x)=1, then 5.f(2) is equal to _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 144
Official Correct Answer:144
Step-by-Step Solution:
f(x)=x6+ax5+bx4+x3
∴f′(x)=6x5+5ax4+4bx3+3x2
Roots 1 & −1
∴6+5z+4b+3=0 & −6+5a−4b+3=0 solving
a=−53
b=−23
∴f(x)=x6−53x5−23x4+x3
∴5.f(2)=5[64−596−24+8]=144
Q 62MathematicsSection A • MCQdifferential equations
+4 / -1
If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is x−2x2−4x+y+8, then this curve also passes through the point :
A
(4, 4)
B
(5, 5)
C
(5, 4)
D
(4, 5)
Official Correct Answer:Option (B)
Step-by-Step Solution:
Given
y (0) = 0
& dxdy=x−2(x−2)2+y+4
⇒dxdy−x−2y=(x−2)+x−24
⇒I.F.=e−∫x−21dx=x−21
Solution of D.E.
⇒y.x−21=∫x−21((x−2)+x−24).dx
⇒x−2y=x−x−24+C
Now, at x = 0, y = 0 ⇒ C = −2
∴ y = x (x − 2) − 4 − 2 (x − 2)
⇒ y = x2− 4x
This curve passes through (5, 5)
Q 63MathematicsSection B • Numericalmatrices and determinants
+4 / -1
If the system of equations
kx + y + 2z = 1
3x − y − 2z = 2
−2x −2y −4z = 3
has infinitely many solutions, then k is equal to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 21
Q 64MathematicsSection B • Numericalvector algebra
+4 / -1
Let a=i+2j−k, b=i−j and c=i−j−k be three given vectors. If r is a vector such that r×a=c×a and r.b=0, then r.a is equal to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 12
Official Correct Answer:12
Step-by-Step Solution:
Given, a=i+2j−k,
b=i−j,
c=i−j−k
r×a=c×a
⇒r×a−c×a=0
⇒(r−c)×a=0
∴r−c=λa
⇒r=λa+c
⇒r.b=λa.b+c.b (taking dot with b)
⇒0=λa.b+c.b [∵r.b=0]
⇒λ(i+2j−k).(i−j)+(i−j−k).(i−j)=0
⇒λ(1−2)+2=0
⇒λ=2
∴r=2a+c
⇒r.a=2a.a+c.a [taking dot with a]
=2a2+a.c
=2(1+4+1)+(1−2+1)
⇒r.a=12
Q 65MathematicsSection A • MCQ3d geometry
+4 / -1
Let α be the angle between the lines whose direction cosines satisfy the equations l + m − n = 0 and l2 + m2− n2 = 0. Then the value of sin4α + cos4α is :
A
83
B
43
C
21
D
85
Official Correct Answer:Option (D)
Step-by-Step Solution:
l2+m2+n2=1
∴2n2=1 (∵ l2 + m2− n2 = 0)
⇒n=±21
∴l2+m2=21 & l+m=21
⇒21−2lm=21
⇒lm=0 or m=0
∴l=0,m=21 or l=21
<0,21,21> or <21,0,21>
∴cosα=0+0+21=21
∴sin4α+cos4α=1−21sin2(2α)=1−21.43=85
Q 66MathematicsSection A • MCQpermutations and combinations
+4 / -1
The total number of positive integral solutions (x, y, z) such that xyz = 24 is :
A
36
B
24
C
45
D
30
Official Correct Answer:Option (D)
Step-by-Step Solution:
x.y.z=24
x.y.z=23.31
Three 2 has to be distributed among x, y and z
Each may receive none, one or two
∴ Number of ways = 3+3−1C3−1 = 5C2 ways
Similarly one 3 has to be distributed among x, y and z
∴ Number of ways = 1+3−1C3−1 = 3C2 ways
Total ways = 5C2.3C2 = 30
Q 67MathematicsSection A • MCQparabola
+4 / -1
A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y = 1. Which of the following points does NOT lie on it?
A
(0, 3)
B
(−6, 0)
C
(4, 5)
D
(5, 4)
Official Correct Answer:Option (D)
Step-by-Step Solution:
Equation of tangent : y=mx+2m3
mT=21 (∵ perpendicular to line 2x+y=1)
∴ tangent is : y=2x+3
⇒x−2y+6=0
Q 68MathematicsSection A • MCQindefinite integration
+4 / -1
The value of the integral ∫1−cos2θsinθ.sin2θ(sin6θ+sin4θ+sin2θ)2sin4θ+3sin2θ+6dθ is :
The equation of the line through the point (0, 1, 2) and perpendicular to the line
2x−1=3y+1=−2z−1 is :
A
3x=−4y−1=3z−2
B
3x=4y−1=−3z−2
C
−3x=4y−1=3z−2
D
3x=4y−1=3z−2
Official Correct Answer:Option (C)
Step-by-Step Solution:
2x−1=3y+1=−2z−1=λ
Any point on this line (2λ+1,3λ−1,−2λ+1)
Direction ratio of given line (2,3,−2)
Direction ratio of line to be found (2λ+1,3λ−2,−2λ−1)
∴d1.d2=0
⇒λ=2/17
Direction ratio of line (21,−28,−21)≡(3,−4,−3)≡(−3,4,3)
Q 77MathematicsSection A • MCQtrigonometric functions and equations
+4 / -1
All possible values of θ∈ [0, 2π] for which sin 2θ + tan 2θ > 0 lie in :
A
(0,4π)∪(2π,43π)∪(23π,611π)
B
(0,2π)∪(π,23π)
C
(0,2π)∪(2π,43π)∪(π,67π)
D
(0,4π)∪(2π,43π)∪(π,45π)∪(23π,47π)
Official Correct Answer:Option (D)
Step-by-Step Solution:
sin2θ+tan2θ>0
⇒sin2θ+cos2θsin2θ>0
⇒sin2θcos2θ(cos2θ+1)>0⇒tan2θ(2cos2θ)>0
Note : cos2θ=0
⇒1−2sin2θ=θ⇒sinθ=±21
Now, tan2θ(1+cos2θ)>0
⇒tan2θ>0 (as cos2θ+1>0)
⇒2θ∈(0,2π)∪(π,23π)∪(2π,25π)∪(3π,27π)
⇒θ∈(0,4π)∪(2π,43π)∪(π,45π)∪(23π,47π)
Q 78MathematicsSection A • MCQdefinite integration
+4 / -1
The value of −1∫1x2e[x3]dx, where [ t ] denotes the greatest integer ≤ t, is :
A
3e+1
B
3ee−1
C
3e1
D
3ee+1
Official Correct Answer:Option (D)
Step-by-Step Solution:
I=−1∫1x2e[x3]dx
Here -1 ≤ x ≤ 1 then -1 ≤ x3≤ 1
Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.
=−1∫0x2e[x3]dx+0∫1x2e[x3]dx
=−1∫0x2e−1dx+0∫1x2e0dx
= e1×[3x3]−10+[3x3]01
=e1×(0−(3−1))+31
=3e1+31=3e1+e
Q 79MathematicsSection B • Numericalsequences and series
+4 / -1
Let A1, A2, A3, ....... be squares such that for each n ≥ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 9
Official Correct Answer:9
Step-by-Step Solution:
∴ Side lengths are in G.P.
Tn=(2)n−112
∴ Area =2n−1144 < 1
⇒2n−1>144
Smallest n = 9
Q 80MathematicsSection B • Numericalhyperbola
+4 / -1
The locus of the point of intersection of the lines (3)kx+ky−43=0 and 3x−y−4(3)k=0 is a conic, whose eccentricity is _________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 2
Official Correct Answer:2
Step-by-Step Solution:
3kx+ky=43 ........(1)
3kx−ky=43k2 ....... (2)
Adding equation (1) & (2)
23kx=43(k2+1)
x=2(k+k1) ......... (3)
Substracting equation (1) & (2)
y=23(k1−k) ........(4)
∴4x2−12y2=4
16x2−48y2=1 (Hyperbola)
∴e2=1+1648
⇒e=2
Q 81MathematicsSection A • MCQcomplex numbers
+4 / -1
Let the lines (2 − i)z = (2 + i)z and (2 + i)z + (i − 2)z− 4i = 0, (here i2 = −1) be normal to a circle C. If the line iz + z + 1 + i = 0 is tangent to this circle C, then its radius is :
A
223
B
32
C
221
D
23
Official Correct Answer:Option (A)
Step-by-Step Solution:
(2−i)z=(2+i)z
⇒(2−i)(x+iy)=(2+i)(x−iy)
⇒2x−ix+2iy+y=2x+ix−2−iy+y
⇒2ix−4iy=0
L1:x−2y=0
⇒(2+i)z+(i−2)z−4i=0
⇒(2+i)(x+iy)+(i−2)(x−iy)−4i=0
⇒2x+ix+2iy−y+ix−2x+y+2iy−4i=0
⇒2ix+4iy−4i=0
L2:x+2y−2=0
Solve L1 and L2: x = 1, 4y=2,y=21
∴x=1
Centre (1,21)
L3:iz+z+1+i=0
⇒i(x+iy)+x−iy+1+i=0
⇒ix−y+x−iy+1+i=0
⇒(x−y+1)+i(x−y+1)=0
Radius = distance from (1,21) to x−y+1=0
⇒r=21−21+1
⇒r=223
Q 82MathematicsSection A • MCQprobability
+4 / -1
The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
A
721
B
2165
C
361
D
541
Official Correct Answer:Option (B)
Step-by-Step Solution:
ax2 + bx + c = 0
a, b, c ∈ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
For equal roots, D = 0 ⇒ b2 = 4ac
⇒ ac = 4b2
Favourable case :
If b = 2, ac = 1 ⇒ a = 1, c = 1
If b = 4, ac = 4 :
a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 ⇒ a = 3, c = 3
∴ Favorable cases = 5
∴ Required probability = 2165
Q 83MathematicsSection A • MCQheight and distance
+4 / -1
A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30∘ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45∘. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
A
10(3+1)
B
10(3−1)
C
10
D
103
Official Correct Answer:Option (A)
Step-by-Step Solution:
x+yh=tan30∘
x+y=3h ...... (1)
Also,
yh=tan45∘
h=y ..... (2)
put in (1)
x+y=3y
x=(3−1)y
20x=′v′ speed
∴ time taken to reach Foot from B
=Vy
=(3−1).xx×20
=10(3+1)
Q 84MathematicsSection B • Numericalarea under the curves
+4 / -1
The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to __________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 64
Official Correct Answer:64
Step-by-Step Solution:
A=4π∫45π(sinx−cosx)dx
=[−cosx−sinx]π/45π/4
=−[(cos45π+sin45π)−(cos4π+sin4π)]
=−[(−21−21)−(21+21)]
=24=22
⇒A4=(22)4=64
Q 85MathematicsSection B • Numericalmatrices and determinants
+4 / -1
If A = \left[ {\matrix{
0 & { - \tan \left( {{\theta \over 2}} \right)} \cr
{\tan \left( {{\theta \over 2}} \right)} & 0 \cr
} } \right] and ({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{
a & { - b} \cr
b & a \cr
} } \right], then 13(a2+b2) is equal to
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 13
Q 86MathematicsSection B • Numericalmatrices and determinants
+4 / -1
Let A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right], where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A2=I3, then the value of x3+y3+z3 is ____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 7
Official Correct Answer:7
Step-by-Step Solution:
A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]
∴∣A∣=(x3+y3+z3−3xyz)
Given A2=I3
∣A2∣=1
∴(x3+y3+z3−3xyz)2=1
⇒x3+y3+z3−3xyz=1 only as (x+y+z>0)
⇒x3+y3+z3=6+1=7
Q 87MathematicsSection A • MCQprobability
+4 / -1
When a missile is fired from a ship, the probability that it is intercepted is 31 and the probability that the missile hits the target, given that it is not intercepted, is 43. If three missiles are fired independently from the ship, then the probability that all three hit the target, is :
A
43
B
83
C
271
D
81
Official Correct Answer:Option (D)
Step-by-Step Solution:
Probability of not getting intercepted = 32
When it is not intercepted, probability of missile hitting target = 43
∴ So when such 3 missiles launched
then P (all 3 hitting the target)
= (32×43)×(32×43)×(32×43)
=81
Q 88MathematicsSection A • MCQfunctions
+4 / -1
Let f, g : N → N such that f(n + 1) = f(n) + f(1) ∀ n∈N and g be any arbitrary function. Which of the following statements is NOT true?
A
If g is onto, then fog is one-one
B
f is one-one
C
If f is onto, then f(n) = n ∀n∈N
D
If fog is one-one, then g is one-one
Official Correct Answer:Option (A)
Step-by-Step Solution:
f(n+1)=f(n)+1
f(2)=2f(1)
f(3)=3f(1)
f(4)=4f(1)
.....
f(n)=nf(1)
f(x) is one-one
Q 89MathematicsSection A • MCQapplication of derivatives
+4 / -1
If Rolle's theorem holds for the function f(x)=x3−ax2+bx−4, x∈[1,2] with f′(34)=0, then ordered pair (a, b) is equal to :
A
(−5, −8)
B
(5, −8)
C
(−5, 8)
D
(5, 8)
Official Correct Answer:Option (D)
Step-by-Step Solution:
f(1)=f(2)
⇒1−a+b−4=8−4a+2b−4
3a−b=7 ..... (1)
f′(x)=3x2−2ax+b
⇒f′(34)=0⇒3×916−38a+b=0
⇒−8a+3b=−16 ..... (2)
∴a=5,b=8
Q 90MathematicsSection B • Numericalpermutations and combinations
+4 / -1
The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________.
Answer Format:Enter numerical value rounded to the nearest integerOfficial Answer: 32
Official Correct Answer:32
Step-by-Step Solution:
The numbers are lying between 100 and 1000
then each number is of three digits.
The possible combination of numbers which are divisible by 3 are 1, 2,
3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.
(If sum of digits of a number is divisible by 3 then the number is divisible by 3)
∴ Total number of numbers = 4 × 3! = 24
The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5;
1, 4, 5 and 2, 4, 5.
(If the last digit of a number is 0 or 5 then the number is divisible by 5)
∴ Total number of numbers = 6 × 2! = 12
The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.