Jee main set B all questions

Chemistry
1. Which of the following is the energy of a possible excited state of hydrogen?
(A) −6.8 eV
(B) −3.4 eV
(C) +6.8 eV
(D) +13.6 eV
Solution: (B) En = −
13.6
n2
eV
Where n = 2 ⇒ E2 = −3.40 eV
Topic: Atomic Structure
Difficulty Level: Easy [Embibe predicted high weightage]
2. In the followingsequence of reactions:
Toluene
KMnO4
→ A
SOCl2
→ B
H2 Pd⁄ ,BaSO4
→ C
The Product C is:
(A) C6H5CH3
(B) C6H5CH2OH
(C) C6H5CHO
(D) C6H5COOH
Solution:(C)
Topic:Aldehyde and Ketone

3. Whichcompoundwouldgive 5-keto-2-methyl hexanal uponozonolysis:
(A)
(B)
(C)
Solution:(A)

Topic: Hydrocarbons
Difficulty Level: Easy [Embibe predicted easy to score (Must Do)]
4. The ionic radii (in Å) of N3−,O2− and F− are respectively:
(A) 1.36, 1.71 and 1.40
(B) 1.71, 1.40 and 1.36
(C) 1.71, 1.36 and 1.40
(D) 1.36, 1.40 and 1.71
Solution: (B) As
Z
e
↑ ionic radius decreases for isoelectronic species.
N3− (
Z
e
) =
7
10
O2− (
Z
e
) =
8
10
F− (
Z
e
) =
9
10
N3− > O2− > F−
Topic: Periodicity
Difficulty Level: Easy [Embibe predicted Low Weightage]
5. The color of KMnO4 isdue to:

(A) d − d transition
(B) L → M charge transfertransition
(C) σ − σ∗transition
(D) M → L charge transfertransition
Solution:(B) Charge transferfrom O2− toemptyd-orbitalsof metal ion (Mn+7)
Topic:Coodination compounds
6. Assertion: Nitrogen and Oxygen are the main components in the atmosphere but these
do not react to form oxides of nitrogen.
Reason: The reaction between nitrogen and oxygen requires high temperature.
(A) Both Assertion and Reason are correct, but the Reason is not the correct explanation for
the Assertion.
(B) The Assertion is incorrect but the Reason is correct.
(C) Both the Assertion and Reason are incorrect.
(D) Both Assertion and Reason are correct and the Reason is the correct explanation for the
Assertion.
Solution: (D) N2(g) & O2(g) react under electric arc at 2000o
Cto form NO(g). Both assertion and
reason are correct and reason is correct explanation.
Topic: Group 15 – 18
Difficulty Level: Moderate [Embibe predicted high weightage]
7. Whichof the followingcompoundsisnotanantacid?
(A) Cimetidine
(B) Phenelzine
(C) Rantidine
(D) Aluminiumhydroxide
Solution:(B) Ranitidine,Cimetidineandmetal hydroxidesi.e.Aluminumhydroxide canbe usedas
antacidbut not phenelzine.Phenelzineisnotan antacid.Itis an antidepressant.Antacidsare atype of

medicationthatcan control the acid levelsinstomach.Workingof antacids:Antacidscounteract
(neutralize) the acidinstomachthat’susedto aiddigestion.Thishelpsreduce the symptomsof
heartburnandrelievespain.
Topic:Chemistry in everyday life
8. In the contextof the Hall-Heroultprocessforthe extractionof Al,whichof the following
statementsisfalse?
(A) Al2O3 is mixedwith CaF2 whichlowersthe meltingpointof the mixture andbringsconductivity.
(B) Al3+ is reducedatthe cathode to formAl.
(C) Na3AlF6 servesas the electrolyte.
(D) CO and CO2 are producedinthisprocess.
Solution:(C) Hall-Heroultprocessforextractionof Al. Al2O3 iselectrolyte Na3AlF6 reducesthe fusion
temperature andprovidesgoodconductivity.
Topic: Metallurgy
9. Match the catalysts to the correct process:
(A) A → ii ,B → i , C → iv ,D → iii
Catalyst Process
A TiCl3 i. Wacker process
B. PdCl2 ii. Ziegler – Natta polymerization
C. CuCl2 iii. Contact process`
D. V2O5 iv. Deacon’s process

(B) A → ii ,B → iii , C → iv , D → i
(C) A → iii ,B → i , C → ii ,D → iv
(D) A → iii ,B → ii , C → iv , D → i
Solution:(A) The Wackerprocessoriginallyreferredtothe oxidationof ethylene toacetaldehyde byoxygenin
waterin the presence of tetrachloropalladate (II) asthe catalyst.
In contact process,Platinumusedtobe the catalystforthisreaction,howeverasitissusceptible toreactingwith
arsenicimpuritiesinthe sulphurfeedstock,vanadium(V) oxide (V2O5)isnow preferred.
In Deacon'sprocess,The reactiontakesplace at about400 to 450oC in the presence of avarietyof catalysts,
includingcopperchloride(CuCl2).
In Ziegler-Nattacatalyst,Homogenouscatalystsusuallybasedoncomplexesof Ti,Zror Hf used.Theyare usually
usedincombinationwithdifferentorgano aluminiumco-catalyst.
Topic: Transition metals
10. In the reaction,
The product E is:
(A)
(B)

(C)
(D)
Solution:(B)
11. Which polymer is used in the manufacture of paints and lacquers?
(A) Glyptal
(B) Polypropene

(C) Poly vinyl chloride
(D) Bakelite
Solution: (A) Glyptal is polymer of glycerol and phthalic anhydride.
Topic:Polymers
12. The number of geometric isomers that can exist for square planar
[Pt(Cl)(py)(NH3)(NH2OH)]+
is (py = pyridine):
(A) 3
(B) 4
(C) 6
(D) 2
Solution:(A) Complexeswithgeneralformula [Mabcd]+ square planarcomplex canhave three
isomers.
Topic:Coordination compounds
13. Higher order (> 3) reactions are rare due to:
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions.
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species

Solution: (D) Probability of an event involving more than three molecules in a collision are remote.
Topic:Chemical kinetics
14. Which among the following is the most reactive?
(A) Br2
(B) I2
(C) ICl
(D) Cl2
Solution: (C) I − Cl bond strength is weaker than I2, Br2 and Cl2 (Homonuclear covalent).
I − Cl bond is polar due to electronegativity difference between I & Cl hence more reactive.
15. Two Faradayof electricityispassedthroughasolutionof CuSO4.The massof copperdeposited
at the cathode is:(Atomicmassof Cu = 63.5 amu)
(A) 63.5 g
(B) 2 g
(C) 127 g
(D) 0 g
Solution:(A) 2F ≡ 2 Eqs of Cu
= 2 ×
63.5
2
= 63.5 g
Topic:Electrochemistry

16. 3 g of activatedcharcoal was addedto 50 mL of acetic acidsolution(0.06N) ina flask.Afteran hour it was
filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per
gram of charcoal) is:
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution: (D) Meqs of CH3COOH (initial) = 50 × 0.06 = 3 Meqs
Meqs CH3COOH (final) = 50 × 0.042 = 2.1 Meqs
CH3 COOH adsorbed = 3 − 2.1 = 0.9 Meqs
= 9 × 10−1 × 60 g Eq⁄ × 10−3 g
= 540 × 10−4 = 0.054 g
= 54 mg
Per gram =
54
3
= 18 mg g⁄ of Charcoal
Topic: Surface chemistry
17. The synthesisof alkyl fluoridesisbestaccomplishedby:
(A) Sandmeyer’s reaction
(B) Finkelsteinreaction
(C) Swartsreaction
(D) Free radical fluorination
Solution:(C) Alkyl fluoridesisbestaccomplishedbyswartsreactioni.e.heatinganalkyl
chloride/bromideinthe presenceof metallicfluoride suchasAgF, Hg2F2,CoF2,SbF3.
CH3 Br + AgF → CH3 − F + AgBr
The reactionof chlorinatedhydrocarbonswithmetallicfluoridestoformchlorofluoro
hydrocarbons,suchas CCl2F2 is knownasswarts reaction.
Topic:Alkyl and Aryl halides

18. The molecular formula of a commercial resin used for exchanging ions in water
softening is C8H7SO3Na (Mol. wt. 206). What would be the maximum uptake of Ca2+
ions by the resin when expressed in mole per gram resin?
(A)
1
206
(B)
2
309
(C)
1
412
(D)
1
103
Solution: (C) 2C8H7 SO3
−
Na+
+ Ca(aq)
2+
→ (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g ≡ 40 g Ca2+
ions
= 1 moles of Ca2+
ions
Moles of Ca2+
g⁄ of resin =
1
412
Topic: s-Block elements and Hydrogen
Difficulty Level: Moderate (Embibe predicted Low Weightage)
19. Whichof the vitaminsgivenbelow iswatersoluble?
(A) Vitamin D
(B) Vitamin E
(C) Vitamin K
(D) Vitamin C
Solution:(D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day.
Topic: Biomolecules

20. The intermolecularinteractionthatis dependentonthe inversecube of distance betweenthe
moleculesis:
(A) Ion-dipoleinteraction
(B) Londonforce
(C) Hydrogenbond
(D) Ion-ioninteraction
Solution:(A) For Ion-dipole interaction
F ∝ μ (
dE
dr
)
μ is dipole moment of dipole
r is distance between dipole and ion.
∝ μ
d
dr
(
1
r2
) ∝
μ
r3
Ion − ion ∝
1
r2
London force ∝
1
r6
Hydrogen bond − Dipole − dipole ∝
1
r4
Topic: Chemical Bonding
21. The following reaction is performed at 298 K.
2NO(g) + O2(g) ⇌ 2NO2(g)
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the
standard free energy of formation of NO2(g) at 298 K? (KP = 1.6 × 1012
)
(A) 86,600 + R(298) ln(1.6 × 1012)
(B) 86,600 −
ln(1.6×1012)
R(298)
(C) 0.5[2 × 86,600 − R(298) ln(1.6 × 1012 )]
(D) R(298) ln(1.6 × 1012) − 86,600

Solution: (C) ΔGreac
o
= −2.303 RTlog KP
= −RT lnKP
= −R(298)ln(1.6 × 1012 )
ΔGreac
o
= 2ΔGf(NO2)
o
− 2ΔGf(NO)g
o
= 2ΔGf(NO2)
o
− 2 × 86.6 × 103
2ΔGf(NO2)
o
= −R(298)ln(1.6 × 1012) + 2 × 86,600
ΔGf(NO2)
o
= 86,600 −
R(298)
2
ln(1.6 × 1012)
= 0.5[2× 86,600 − R(298)ln(1.6× 1012 )]
Topic: Thermochemistry and thermodynamics
22. Whichof the followingcompoundsisnotcoloredyellow?
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution:(D) Cyanides notyellow.
BaCrO4 − Yellow
K3[Co(NO2)6]− Yellow
(NH4)3[As(Mo3O10)4] − Yellow
Zn2[Fe(CN)6]is bluishwhite.
Topic:Salt Analysis
23. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141
mg of AgBr. The percentage of bromine in the compound is : (Atomic mass : Ag = 108, Br
= 80)
(A) 36

(B) 48
(C) 60
(D) 24
Solution: (D) R − Br
Carius method
→ AgBr
250 mg organic compound is RBr
141 mg AgBr ⇒ 141 ×
80
188
mg Br
% Br in organic compound
= 141 ×
80
188
×
1
250
× 100 = 24%
Topic: Stoichiometry
24. Sodiummetal crystallizesinabodycentredcubiclattice withaunitcell edge of 4.29 Å . The
radiusof sodiumatomisapproximately:
(A) 3.22 Å
(B) 5.72 Å
(C) 0.93 Å
(D) 1.86 Å
Solution:(D)
For B.C.C
4r = √3a
r =
√3
4
a =
1.732
4
× 4.29
= 1.86 Å
Topic: Solid State
25. Which of the following compounds will exhibit geometrical isomerism?

(A) 3 – Phenyl – 1- butene
(B) 2 – Phenyl – 1- butene
(C) 1,1 – Diphenyl – 1- propane
(D) 1 – Phenyl – 2 – butene
Solution: (D) 1 - Phenyl - 2 - butene:
Topic:General Organic Chemistry and Isomerism
26. The vapourpressure of acetone at 20oC is185 torr. When1.2 g of a non-volatile substance wasdissolvedin
100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol−1)of the substance is :
(A) 64
(B) 128
(C) 488
(D) 32
Solution: (A) ΔP = 185 − 183 = 2 torr
ΔP
Po =
2
185
= XB =
1.2
M
(
1.2
M
) +
100
58
M(CH3)2CO = 15 × 2 + 16 + 12 = 58 g mole⁄
1.2
M
≪
100
58
⇒
2
185
=
1.2
M
×
58
100
M =
58 × 1.2
100
×
185
2

= 64.38 ≈ 64 g mole⁄
Topic: Liquid solution
27. From the following statements regarding H2O2 , choose the incorrect statement:
(A) It decomposesonexposure tolight
(B) It has to be storedinplasticor wax linedglassbottlesindark.
(C) It has to be keptawayfrom dust.
(D) It can act onlyas an oxidizingagent.
Solution: (D) It can act both as oxidizing agent and reducing agent.
Topic: s-Block elements and hydrogen
28. Which one of the following alkaline earth metal sulphates has its hydration enthalpy
greater than its lattice enthalpy?
(A) BeSO4
(B) BaSO4
(C) SrSO4
(D) CaSO4
Solution: (A) ΔHHydration > ΔHLattice
Salt is soluble. BeSO4 is soluble due to high hydration energy of small Be2+
ion. Ksp for
BeSO4 is very high.
Topic: s-Block elements and hydrogen
29. The standard Gibbsenergychange at 300 K for the reaction 2A ⇌ B + C is2494.2 J.At a given
time,the compositionof the reactionmixture is [A] =
1
2
, [B] = 2 and[C] =
1
2
. The reaction
proceedsinthe: [R = 8.314 J K − mol , e = 2.718]⁄
(A) Reverse direction because Q > KC

(B) Forward direction because Q < KC
(C) Reverse direction because Q < KC
(D) Forward direction because Q > KC
Solution:(A) ∆Go = −RT In KC
2494.2 = −8.314 × 300 in KC
2494.2 = −8.314 × 300 × 2.303log KC
−
2494.2
2.303×300×8.314
= −0.44 = log KC
logKC = −0.44 = 1̅. 56
KC = 0.36
QC =
2×
1
2
(1
2
)
2 = 4
QC > KC reverse direction.
Topic: Chemical Equilibrium
Difficulty Level: Moderate [Embibe predicted easy to score (Must Do)]
30. Whichone has the highestboilingpoint?
(A) Ne
(B) Kr
(C) Xe
(D) He
Solution:(C) Due tohigherVanderWaal’sforces.Xe hasthe highestboilingpoint.

Mathematics
31. The sum of coefficients of integral powers of x in the binomial expansion of (1 − 2√ 𝑥)
50
is:
(A)
1
2
(350)
(B)
1
2
(350
− 1)
(C)
1
2
(250
+ 1)
(D)
1
2
(350
+ 1)
Answer: (D)
Solution:
Integral powers ⇒ odd terms
𝑜𝑑𝑑 𝑡𝑒𝑟𝑚𝑠 =
(1 − 2√ 𝑥 )
50
+ (1 + 2√ 𝑥 )
50
2
𝑆𝑢𝑚 𝑜𝑓 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 =
(1 − 2)50
+ (1 + 2)50
2
=
1 + 350
2
Topic: Binomial Theorem
Difficulty level: Moderate (embibe predicted high weightage)
32. Let 𝑓( 𝑥) be a polynomial of degree four having extreme values at 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 2. If
𝑙𝑖𝑚
𝑥 → 0
[1 +
𝑓(𝑥)
𝑥2 ] = 3, 𝑡ℎ𝑒𝑛 𝑓(2) is equal to:
(A) −4
(B) 0

(C) 4
(D) −8
Answer: (B)
Solution:
Let 𝑓( 𝑥) = 𝑎𝑥4
+ 𝑏𝑥3
+ 𝑐𝑥2
+ 𝑑𝑥 + 𝑒
𝑙𝑖𝑚
𝑥 ⟶ 0
[1 +
𝑓(𝑥)
𝑥2 ] = 3
𝑙𝑖𝑚
𝑥 ⟶ 0
[1 + 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 +
𝑑
𝑥
+
𝑒
𝑥2 ] = 3
This limit exists when d = e = 0
So,
𝑙𝑖𝑚
𝑥 ⟶ 0
[1 + 𝑎𝑥2
+ 𝑏𝑥 + 𝑐] = 3
⇒ 𝑐 + 1 = 3
𝑐 = 2
It is given, 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 2 are solutions of 𝑓′( 𝑥) = 0
𝑓′( 𝑥) = 4𝑎𝑥3
+ 3𝑏𝑥2
+ 2𝑐𝑥
𝑥(4𝑎𝑥2
+ 3𝑏𝑥 + 2𝑐) = 0
1,2 are roots of quadratic equation
⇒ 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑜𝑡𝑠 =
− 3𝑏
4 𝑎
= 1 + 2 = 3
⇒ 𝑏 = −4𝑎
Product of roots =
2𝑐
4𝑎
= 1.2 = 2

⇒ 𝑎 =
𝑐
4
𝑎 =
1
2
, 𝑏 = −2
∴ 𝑓( 𝑥) =
1
2
𝑥4
− 2𝑥3
+ 2𝑥2
𝑓 (2) = 8 − 16 + 8
= 0
Topic: Application of Derivatives
Difficulty level: Moderate ( embibe predicted high weightage)
33. The mean of the data set comprising of 16 observations is 16. If one of the observation
valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then
the mean of the resultant data, is :
(A) 16.0
(B) 15.8
(C) 14.0
(D) 16.8
Answer: (C)
Solution:
Given,
∑ 𝑥𝑖+1615
𝑖=1
16
= 16
⇒ ∑ 𝑥𝑖 + 16 = 256
15
𝑖=1
∑ 𝑥𝑖 = 240
15
𝑖=1

Required mean =
∑ 𝑥𝑖+3+4+515
𝑖=1
18
=
240 +3+4+5
18
=
252
18
= 14
Topic: Statistics
Difficulty level: Moderate (embibe predicted easy to score (Must Do))
34. The sum of first 9 terms of the series
13
1
+
13
+23
1+3
+
13
+23
+33
1+3+5
+ ⋯ 𝑖𝑠:
(A) 96
(B) 142
(C) 192
(D) 71
Answer: (A)
Solution:
𝑇𝑛 =
13
+23
+⋯ +𝑛3
1+3+⋯+(2𝑛−1)
=
𝑛2 ( 𝑛+1)2
4
𝑛2 =
( 𝑛+1)2
4
Sum of 9 terms = ∑
( 𝑛+1)2
4
9
𝑛=1
=
1
4
× [22
+ 32
+ ⋯ + 102]
=
1
4
[(12
+ 22
+ ⋯ + 102) − 12]
=
1
4
[
10.11.21
6
− 1]
=
1
4
[384] = 96
Topic: Progression & Series
35. Let O be the vertex and Q be any point on the parabola, 𝑥2
= 8𝑦. If the point P divides the
line segment OQ internally in the ratio 1 : 3, then the locus of P is:

(A) 𝑦2
= 𝑥
(B) 𝑦2
= 2𝑥
(C) 𝑥2
= 2𝑦
(D) 𝑥2
= 𝑦
Answer: (C)
Solution:
General point on 𝑥2
= 8𝑦 is 𝑄 (4𝑡, 2𝑡2)
Let P(h, k) divide OQ in ratio 1:3
(h, k) = (
1 (4𝑡)+3(0)
1+3
,
1(2𝑡2 )+3(0)
1+3
)
(h, k) = (𝑡,
2𝑡2
4
)
h = t and 𝑘 =
𝑡2
2
𝑘 =
ℎ2
2
⇒ 𝑥2
= 2𝑦 is required locus.

Topic: Parabola
36. Let 𝛼 𝑎𝑛𝑑 𝛽 be the roots of equation𝑥2
− 6𝑥 − 2 = 0. If 𝑎 𝑛 = 𝛼 𝑛
− 𝛽 𝑛
, for 𝑛 ≥ 1, then the
value of
𝑎10 −2𝑎8
2𝑎9
is equal to :
(A) -6
(B) 3
(C) -3
(D) 6
Answer: (B)
Solution:
Given, 𝛼, 𝛽 are roots of 𝑥2
− 6𝑥 − 2 = 0
⇒ 𝛼2
− 6𝛼 − 2 = 0 and 𝛽2
− 6𝛽 − 2 = 0
⇒ 𝛼2
− 6 = 6𝛼 and 𝛽2
− 2 = 6𝛽 …..(1)
𝑎10 − 2𝑎8
2𝑎9
=
( 𝛼10
− 𝛽10) − 2(𝛼8
− 𝛽8
)
2(𝛼9 − 𝛽9)
=
( 𝛼10
−2𝛼8 )−(𝛽10
−2𝛽8
)
2(𝛼9−𝛽9)
=
𝛼8 ( 𝛼2
−2)−𝛽8
(𝛽2
−2)
2(𝛼9 −𝛽9)
=
𝛼8 (6𝛼)−𝛽8
(6𝛽)
2(𝛼9 −𝛽9)
[∵ 𝐹𝑟𝑜𝑚 (1)]
=
6𝛼9
−6𝛽9
2(𝛼9 −𝛽9)
= 3
Topic: Quadratic Equation & Expression
37. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the
boxes contains exactly 3 balls is:

(A) 55(
2
3
)
10
(B) 220(
1
3
)
12
(C) 22(
1
3
)
11
(D)
55
3
(
2
3
)
11
Answer: (C)
Solution: (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0

7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19.
Number of favorable case = 5
Required probability =
5
19
INCORRECT SOLUTION
Required Probability = 12
𝐶9 ⋅ (
1
3
)
3
⋅ (
2
3
)
9
= 55 ⋅ (
2
3
)
11
The mistake is “we can’t use 𝑛
𝐶𝑟 for identical objects”.
Topic: Probability
Difficulty level: Difficult (embibe predicted high weightage)

38. A complex number z is said to be unimodular if | 𝑧| = 1. suppose 𝑧1 𝑎𝑛𝑑 𝑧2 are complex
numbers such that
𝑧1−2𝑧2
2−𝑧1 𝑧̅2
is unimodular and 𝑧2 is not unimodular. Then the point 𝑧1lies on a:
(A) Straight line parallel to y-axis.
(B) Circle of radius 2
(C) Circle of radius √2
(D) Straight line parallel to x-axis.
Answer: (B)
Solution:
Given,
𝑧1− 2𝑍2
2−𝑧1 𝑧̅2
is unimodular
⇒ |
𝑧1− 2𝑍2
2−𝑧1 𝑧̅2
| = 1
|𝑧1− 2𝑍2
| = |2 − 𝑧1 𝑧̅2|
Squaring on both sides.
|𝑧1− 2𝑍2
|
2
= |2 − 𝑧1 𝑧̅2|2
(𝑧1−2𝑍2
)( 𝑧1̅ − 2𝑧̅2) = (2 − 𝑧1 𝑧̅2)(2 − 𝑧1̅ 𝑧2)

Topic: Complex Number
Difficulty level: Easy (embibe predicted high weightage)
39. The integral ∫
𝑑𝑥
𝑥2 ( 𝑥4+1)
3
4
equals:
(A) ( 𝑥4
+ 1)
1
4 + 𝑐
(B) −( 𝑥4
+ 1)
1
4 + 𝑐
(C) −(
𝑥4
+1
𝑥4 )
1
4
+ 𝑐
(D) (
𝑥4
+1
𝑥4 )
1
4
+ 𝑐
Answer: (A)
Solution:
𝐼 = ∫
𝑑𝑥
𝑥2( 𝑥4+1)
3
4

= ∫
𝑑𝑥
𝑥2 × 𝑥3 (1+
1
𝑥4)
3
4
𝑃𝑢𝑡 1 +
1
𝑥4 = 𝑡
⇒
−4
𝑥5 𝑑𝑥 = 𝑑𝑡
∴ 𝐼 = ∫
−
𝑑𝑡
4
𝑡
3
4
= −
1
4
× ∫ 𝑡
−3
4 𝑑𝑡
=
−1
4
× (
𝑡
1
4
1
4
) + 𝑐
= − (1 +
1
𝑥4)
1
4
+ c
Topic: Indefinite Integral
Difficulty level: Moderate (embibe predicted Low Weightage)
40. The number of points, having both co-ordinates as integers, that lie in the interior of the
triangle with vertices (0, 0), (0, 41) and (41, 0), is:
(A) 861
(B) 820
(C) 780
(D) 901
Answer: (C)

Solution:
𝑃( 𝑥1, 𝑦1) lies inside the triangle ⇒ 𝑥1, 𝑦1 𝜖 𝑁
𝑥1 + 𝑦1 < 41
∴ 2 ≤ 𝑥1 + 𝑦1 ≤ 40
Number of points inside = Number of solutions of the equation
𝑥1 + 𝑦1 = 𝑛
2 ≤ 𝑛 ≤ 40
𝑥1 ≥ 1, 𝑦1 ≥ 1
( 𝑥1 − 1) + ( 𝑦1 − 1) = ( 𝑛 − 2)
(𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑛 − 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑥1 + 𝑥2 + ⋯. +𝑥 𝑛 = 𝑟 𝑖𝑠 𝑛+𝑟−1
𝐶 𝑟 )
Number of solutions = 2+( 𝑛−2)−1
𝐶 𝑛−2
= 𝑛−1
𝐶 𝑛−2
= 𝑛 − 1
We have 2 ≤ 𝑛 ≤ 40
∴ Number of solutions = 1 + 2 +….+39
=
39 ×40
2
= 780.
Topic: Straight lines
41. The distance of the point (1, 0, 2) from the point of intersection of the line
𝑥−2
3
=
𝑦+1
4
=
𝑧−2
12
and the plane 𝑥 − 𝑦 + 𝑧 = 16, is

(A) 8
(B) 3√21
(C) 13
(D) 2√14
Answer: (C)
Solution:
𝑥−2
3
=
𝑦+1
4
=
𝑧−1
12
= 𝑡 (𝑠𝑎𝑦)
⇒ General point is (2 + 3𝑡, −1 + 4𝑡, 2 + 12𝑡)
It lies on the plane 𝑥 − 𝑦 + 𝑧 = 16
⇒ 2 + 3𝑡 + 1 − 4𝑡 + 2 + 12𝑡 = 16
⇒ 11𝑡 = 11
⇒ 𝑡 = 1
∴ The point of intersection will be
(2 + 3(1),−1 + 4(1),2 + 12(1))
= (5, 3, 14)
Distance from (1,0,2) = √(5 − 1)2 + (3 − 0)2 + (14 − 2)2
= √42 + 32 + 122
= 13
Topic: 3 - D Geometry

42. The equation of the plane containing the line 2𝑥 − 5𝑦 + 𝑧 = 3 ; 𝑥 + 𝑦 + 4𝑧 = 5, and
parallel to the plane, 𝑥 + 3𝑦 + 6𝑧 = 1, is:
(A) 𝑥 + 3𝑦 + 6𝑧 = −7
(B) 𝑥 + 3𝑦 + 6𝑧 = 7
(C) 2𝑥 + 6𝑦 + 12𝑧 = −13
(D) 2𝑥 + 6𝑦 + 12𝑧 = 13
Answer: (B)
Solution:
Any plane containing the line of intersection of 2𝑥 − 5𝑦 + 𝑧 = 3, 𝑥 + 𝑦 + 4𝑧 = 5 will be of
the form
(2𝑥 − 5𝑦 + 𝑧 − 3) + 𝜆 ( 𝑥 + 𝑦 + 4𝑧 − 5) = 0
(2 + 𝜆) 𝑥 − (5 − 𝜆) 𝑦 + (1 + 4𝜆) 𝑧 − (3 + 5𝜆) = 0
(2 + 𝜆) 𝑥 − (5 − 𝜆) 𝑦 + (1 + 4𝜆) 𝑧 − (3 + 5𝜆) = 0
It is parallel to plane 𝑥 + 3𝑦 + 6𝑧 = 1
⇒
2 + 𝜆
1
=
𝜆 − 5
3
=
1 + 4𝜆
6
≠
−(3 + 5𝜆)
11
2 + 𝜆
1
=
𝜆 − 5
3
⇒ 6 + 3𝜆 = 𝜆 − 5
2𝜆 = −11
𝜆 =
−11
2
∴ Required plane is

(2𝑥 − 5𝑦 + 𝑧 − 3)
−11
2
( 𝑥 + 𝑦 + 4𝑧 − 5) = 0
2(2𝑥 − 5𝑦 + 𝑧 − 3) − 11 ( 𝑥 + 𝑦 + 4𝑧 − 5) = 0
4𝑥 − 10𝑦 + 2𝑧 − 6 − 11𝑥 − 11𝑦 − 44𝑧 + 55 = 0
−7𝑥 − 21𝑦 − 42𝑧 + 49 = 0
⇒ 𝑥 + 3𝑦 + 6𝑧 − 7 = 0
Topic: 3 - D Geometry
43. The area (in sq. units) of the region described by [( 𝑥, 𝑦): 𝑦2
≤ 2𝑥 and 𝑦 ≥ 4𝑥 − 1] is
(A)
5
64
(B)
15
64
(C)
9
32
(D)
7
32
Answer: (C)
Solution
Let us find the points intersections of 𝑦2
= 2𝑥 and 𝑦 = 4𝑥 − 1
(4𝑥 − 1)2
= 2𝑥
16𝑥2
− 10𝑥 + 1 = 0
(8𝑥 − 1) (2𝑥 − 1) = 0
𝑥 =
1
2
,
1
8

𝑥 =
1
2
⇒ 𝑦 = 4 (
1
2
)− 1 = 1
𝑥 =
1
8
⇒ 𝑦 = 4 (
1
8
)− 1 =
−1
2
Required area = ∫ (𝑥
1
−1
2 2
− 𝑥1) 𝑑𝑦
= ∫ [
𝑦2
2
− (
𝑦 + 1
4
)]
1
−1
2
𝑑𝑦
=
1
6
× ⌊ 𝑦3⌋−1
2
1
−
1
8
⌊ 𝑦2⌋−1
2
1
−
1
4
⌊ 𝑦⌋−1
2
1
=
1
6
[1 +
1
8
] −
1
8
[1 −
1
4
] −
1
4
[1 +
1
2
]
=
1
6
×
9
8
−
1
8
×
3
4
−
1
4
×
3
2
=
3
16
−
3
32
−
3
8
=
6 − 3 − 12
32
=
−9
32
sq.units
Topic: Area under Curves

44. If m is the A.M. of two distinct real numbers 𝑙 𝑎𝑛𝑑 𝑛 ( 𝑙, 𝑛 > 1) 𝑎𝑛𝑑 𝐺1, 𝐺2 𝑎𝑛𝑑 𝐺3 are three
geometric means between 𝑙 𝑎𝑛𝑑 𝑛, 𝑡ℎ𝑒𝑛 𝐺1
4
+ 2𝐺2
4
+ 𝐺3
4
equals.
(A) 4 𝑙𝑚2
𝑛
(B) 4 𝑙𝑚𝑛2
(C) 4 𝑙2
𝑚2
𝑛2
(D) 4 𝑙2
𝑚𝑛
Answer: (A)
Solution:
m is the A.M. of 𝑙, 𝑛
⇒ 𝑚 =
𝑙+𝑛
2
….(1)
𝐺1, 𝐺2, 𝐺3 are G.M. of between l and n
⇒ 𝑙, 𝐺1, 𝐺2, 𝐺3, 𝑛 are in G.P.
𝑛 = 𝑙( 𝑟)4
⇒ 𝑟4
=
𝑛
𝑙
….(2)
𝐺1 = 𝑙𝑟, 𝐺2 = 𝑙𝑟2
, 𝐺3 = 𝑙𝑟3
𝐺1
4
+ 2𝐺2
4
+ 𝐺3
4
= 𝑙4
𝑟4
+ 2 × 𝑙4
𝑟8
+ 𝑙4
× 𝑟12
= 𝑙4
𝑟4[1 + 2𝑟4
+ 𝑟8]
= 𝑙4
×
𝑛
𝑙
[1 + 𝑟4 ]2
= 𝑙3
𝑛 × [1 +
𝑛
𝑙
]
2
= 𝑙3
× 𝑛 ×
( 𝑙+𝑛)2
𝑙2

= 𝑙𝑛 × (2𝑚)2 [∵ 𝑓𝑟𝑜𝑚 (1)]
= 4𝑙𝑚2
𝑛
Topic: Progression & Series
45. Locus of the image of the point (2, 3) in the line (2𝑥 − 3𝑦 + 4) + 𝑘 ( 𝑥 − 2𝑦 + 3) = 0, 𝑘 ∈
𝑅, is a :
(A) Straight line parallel to y-axis.
(B) Circle of radius √2
(C) Circle of radius √3
(D) Straight line parallel to x-axis
Answer: (B)
Solution:
Given, family of lines 𝐿1 + 𝐿2 = 0
Let us take the lines to be
𝐿2 + 𝜆( 𝐿1 − 𝐿2) = 0
( 𝑥 − 2𝑦 + 3) + 𝜆( 𝑥 − 𝑦 + 1) = 0
(1 + 𝜆) 𝑥 − (2 + 𝜆) 𝑦 + (3 + 𝜆) = 0
Let, Image of (2, 3) be (h, k)
ℎ − 2
1 + 𝜆
=
𝑘 − 3
−(2 + 𝜆)
=
−2(2+ 2𝜆 − 6 − 3𝜆 + 3 + 𝜆)
(1 + 𝜆)2 + (2 + 𝜆)2
ℎ−2
𝜆+1
=
𝑘−3
−(𝜆+2)
=
−2(−1)
(1+𝜆)2+(2+𝜆)2 (1)
ℎ − 2
𝜆 + 1
=
𝑘 − 3
−(𝜆 + 2)
⇒
𝜆 + 2
𝜆 + 1
=
3 − 𝑘
ℎ − 2

1 +
1
𝜆+1
=
3−𝑘
ℎ−2
1
𝜆+1
=
5−ℎ−𝑘
ℎ−2
(2)
From (1): (ℎ − 2)2
+ ( 𝑘 − 3)2
=
4
( 𝜆+1)2+( 𝜆+2)2 (3)
From (1) and (3):
ℎ − 2
𝜆 + 1
=
2
( 𝜆 + 1)2 + ( 𝜆 + 2)2
5 − ℎ − 𝑘 =
1
2
[(ℎ − 2)2
+ ( 𝑘 − 3)2]
10 − 2ℎ − 2𝑘 = ℎ2
+ 𝑘2
− 4ℎ − 6𝑘 + 13
𝑥2
+ 𝑦2
− 2𝑥 − 4𝑦 + 3 = 0
Radius = √1 + 4 − 3
= √2.
Topic: Straight lines
Difficulty level: Difficult (embibe predicted high weightage)
46. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse
𝑥2
9
+
𝑦2
5
= 1, is :
(A) 18
(B)
27
2
(C) 27
(D)
27
4
Answer: (C)
Solution:

𝑥2
9
+
𝑦2
5
= 1
𝑒 = √1 −
5
9
=
2
3
𝑎2
= 9, 𝑏2
= 5
Focii = (± 𝑎𝑒,0) = (± 2,0)
Ends of latus recta = (± 𝑎𝑒, ±
𝑏2
𝑎
)
= (± 2,±
5
3
)
Tangent at ‘L’ is T = 0
2 ⋅ 𝑥
9
+
5
3
⋅
𝑦
5
= 1
It cut coordinates axes at 𝑃 (
9
2
, 0) 𝑎𝑛𝑑 𝑄(0,3)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (
1
2
⋅
9
2
⋅ 3) = 27 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠.
Topic: Ellipse
47. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and
8, without repetition, is:
(A) 192

(B) 120
(C) 72
(D) 216
Answer: (A)
Solution:
4 digit and 5 digit numbers are possible
4 digit numbers:
6/7/8
↑̅ ↑̅ ↑̅
3 4 3
↑̅
2
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers:
↑̅ ↑̅ ↑̅
5 4 3
↑̅ ↑̅
2 1
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
∴ 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 72 + 120 = 192.
Topic: Binomial Theorem
48. Let A and B be two sets containing four and two elements respectively. Then the number of
subsets of the set 𝐴 × 𝐵, each having at least three elements is:
(A) 256
(B) 275
(C) 510
(D) 219

Answer: (D)
Solution:
𝑛( 𝐴) = 4
𝑛( 𝐵) = 2
Number of elements in 𝐴 × 𝐵 = 2 ⋅ 4 = 8
Number of subsets having at least 3 elements
= 28
−8
𝐶0 − 𝐶8
1 − 𝐶8
2
= 256 − 1 − 8 − 28
= 219
Topic: Set and Relations
Difficulty level: Moderate(embibe predicted high weightage)
49. Let tan−1
𝑦 = tan−1
𝑥 + tan−1
(
2𝑥
1−𝑥2 ), where | 𝑥| <
1
√3
, Then a value of y is:
(A)
3𝑥+𝑥3
1−3𝑥2
(B)
3𝑥−𝑥2
1+3𝑥2
(C)
3𝑥+𝑥3
1+3𝑥2
(D)
3𝑥−𝑥3
1−3𝑥2
Answer: (D)
Solution:
tan−1
𝑦 = tan−1
𝑥 + tan−1
(
2𝑥
1−𝑥2), | 𝑥| <
1
√3
= tan−1
𝑥 + 2 tan−1
𝑥
= 3tan−1
𝑥
tan−1
𝑦 = tan−1
(
3𝑥−𝑥3
1−3𝑥2)

⇒ 𝑦 =
3𝑥−𝑥3
1−3𝑥2
Topic: Inverse Trigonometric Function
50. The integral ∫
log 𝑥2
log 𝑥2 +log (36 − 12𝑥 + 𝑥2 )
𝑑𝑥
4
2
is equal to :
(A) 4
(B) 1
(C) 6
(D) 2
Answer: (B)
Solution:
𝐼 = ∫
log 𝑥2
log 𝑥2 + log(6 − 𝑥)2
4
2
𝑑𝑥
𝐼 = ∫
log(6 − 𝑥)2
log(6 − 𝑥)2 + log 𝑥2
4
2
𝑑𝑥
[∵ ∫ 𝑓( 𝑥) 𝑑𝑥 = ∫ 𝑓( 𝑎 + 𝑏 − 𝑥) 𝑑𝑥
𝑏
𝑎
𝑏
𝑎
]
2𝐼 = ∫
log 𝑥2
+ log(6 − 𝑥)2
4
2
𝑑𝑥
2𝐼 = ∫
log 𝑥2
+ log(6 − 𝑥)2
𝑑𝑥
4
2
2𝐼 = ∫1 𝑑𝑥
4
2
2𝐼 = [ 𝑥]2
4

2𝐼 = 4 − 2
𝐼 = 1.
Topic: Definite Integral
51. The negation of ~𝑠 ∨ (~𝑟 ∧ 𝑠) is equivalent to:
(A) 𝑠 ∧ ( 𝑟 ∧ ∼ 𝑠)
(B) 𝑠 ∨ ( 𝑟 ∨ ∼ 𝑠)
(C) 𝑠 ∧ 𝑟
(D) 𝑠 ∧∼ 𝑟
Answer: (C)
Solution:
~ ((~𝑠) ∨ (∼ 𝑟 ∧ 𝑠)) = 𝑠 ∧ [∼ ((∼ 𝑟) ∧ 𝑠)] [∵ ~( 𝑝 ∧ 𝑞) = (∼ 𝑝) ∨ (∼ 𝑞)]
= 𝑠 ∧ (𝑟 ∨ (∼ 𝑠)) [∼ ( 𝑝 ∨ 𝑞) = (∼ 𝑝) ∧ (∼ 𝑞)]
= ( 𝑠 ∧ 𝑟) ∨ (𝑠 ∧ (∼ 𝑠))
= ( 𝑠 ∧ 𝑟) ∨ 𝐹
= 𝑠 ∧ 𝑟 (𝐹 → 𝐹𝑎𝑙𝑙𝑎𝑐𝑦)
Topic: Mathematical Reasoning
52. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a
line leading to the foot of the tower, are 30°, 45° 𝑎𝑛𝑑 60° respectively, then the ratio, 𝐴𝐵 ∶
𝐵𝐶, is:
(A) √3 ∶ √2
(B) 1 ∶ √3

(C) 2 ∶ 3
(D) √3 ∶ 1
Answer: (D)
Solution:
tan 30 𝑜
=
ℎ
𝑥 + 𝑦 + 𝑧
,tan 450
=
ℎ
𝑦 + 𝑧
, tan600
=
ℎ
𝑧
⇒ 𝑥 + 𝑦 + 𝑧 = √3 ℎ
𝑦 + 𝑧 = ℎ
𝑧 =
ℎ
√3
𝑦 = ℎ (1 −
1
√3
)
𝑥 = (√3 − 1)ℎ
𝑥
𝑦
=
(√3 − 1)ℎ
ℎ(
√3 − 1
√3
)
=
√3
1
Topic: Height & Distance
53.
𝑙𝑖𝑚
𝑥 → 0
(1−cos 2𝑥) (3+cos 𝑥)
𝑥 tan 4𝑥
is equal to:
(A) 3

(B) 2
(C)
1
2
(D) 4
Answer: (B)
Solutions:
lim
𝑥→𝑜
(1 − cos2𝑥) (3+ cos 𝑥)
𝑥 tan 4𝑥
= lim
𝑥→𝑜
2 sin2
𝑥 ⋅ (3 + cos 𝑥)
𝑥 ⋅ tan 4𝑥
= lim
𝑥→𝑜
2 ⋅ (
sin 𝑥
𝑥
)
2
⋅ (3 + cos 𝑥)
tan 4𝑥
4𝑥
⋅ 4
=
2 ⋅ (1)2
⋅ (3 + 1)
1 ⋅ 4
= 2.
Topic: Limit,Continuity & Differentiability
54. Let 𝑎⃗, 𝑏⃗⃗, 𝑐⃗ be three non-zero vectors such that no two of them are collinear and (𝑎⃗ × 𝑏⃗⃗) × 𝑐⃗
=
1
3
|𝑏⃗⃗| | 𝑐⃗| | 𝑎⃗|. If 𝜃 is the angle between vectors 𝑏⃗⃗ 𝑎𝑛𝑑 𝑐⃗, then a value of sin 𝜃 is:
(A)
−√2
3
(B)
2
3
(C)
−2√3
3
(D)
2√2
3
Answer: (D)

Solution:
(𝑎⃗ × 𝑏⃗⃗) × 𝑐⃗ =
1
3
|𝑏⃗⃗| | 𝑐⃗| 𝑎⃗
( 𝑎⃗ ⋅ 𝑐⃗) 𝑏⃗⃗− (𝑏⃗⃗ ⋅ 𝑐⃗) 𝑎⃗ =
1
3
|𝑏⃗⃗| | 𝑐⃗| 𝑎⃗
⇒ 𝑎⃗ ⋅ 𝑐⃗ = 0 𝑎𝑛𝑑 𝑏⃗⃗ ⋅ 𝑐⃗ =
−1
3
|𝑏⃗⃗| | 𝑐⃗|
|𝑏⃗⃗| | 𝑐⃗| 𝑐𝑜𝑠𝜃 =
−1
3
|𝑏⃗⃗| | 𝑐⃗|
𝑐𝑜𝑠𝜃 =
−1
3
𝑠𝑖𝑛𝜃 = √1 −
1
9
= √
8
9
=
2√2
3
Topic: Vector Algebra
55. If 𝐴 = [
1 2 2
2 1 −2
𝑎 2 𝑏
] is a matrix satisfying the equation 𝐴𝐴 𝑇
= 9𝐼, where 𝐼 𝑖𝑠 3 × 3 identity
matrix, then the ordered pairs (a, b) is equal to:
(A) (−2,1)
(B) (2,1)
(C) (−2,−1)
(D) (2, −1)
Answer: (C)
Solution:
𝐴𝐴 𝑇
= 9𝐼
[
1 2 2
2 1 −2
𝑎 2 𝑏
] [
1 2 𝑎
2 1 2
2 −2 𝑏
] = [
9 0 0
0 9 0
0 0 9
]

[
1(1) + 2(2)
2(1) + 1(2)
𝑎(1) + 2(2)
+ 2(2)
− 2(2)
+ 𝑏(2)
1(2) + 2(1)
2(2) + 1(1)
𝑎(2) + 2(1)
+ 2(−2)
− 2(−2)
+ 𝑏(−2)
1(𝑎) +
2(𝑎) +
𝑎(𝑎) +
2(2) + 2(𝑏)
1(2) − 2(𝑏)
2(2) + 𝑏(𝑏)
]
= [
9 0 0
0 9 0
0 0 9
]
[
9 0 𝑎 + 2𝑏 + 4
0 9 2𝑎 − 2𝑏 + 2
𝑎 + 2𝑏 + 4 2𝑎 − 2𝑏 + 2 𝑎2
+ 𝑏2
+ 4
] = [
9 0 0
0 9 0
0 0 9
]
𝐶𝑜𝑚𝑝𝑎𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝑎 + 2𝑏 +4 = 0
2𝑎 − 2𝑏 +2 = 0
and 𝑎2
+ 𝑏2
+ 4 = 9
____________________
3𝑎 + 6 = 0
𝑎 = −2 ⇒ −2 + 2𝑏 + 4 = 0
𝑏 = −1
The third equation is useful to verify whether this multiplication is possible.
Topic: Matrices
56. If the function. 𝑔( 𝑥) = {
𝑘 √ 𝑥 + 1, 0 ≤ 𝑥 ≤ 3
𝑚𝑥 + 2, 3 < 𝑥 ≤ 5
is differentiable, then the value of k + m is :
(A)
16
5
(B)
10
3
(C) 4

(D) 2
Answer: (D)
Solution:
𝑔(𝑥) is differentiable at x = 3
⇒ 𝑔 (𝑥) is continuous at x = 3
∴ LHL = RHL = 𝑔(3)
lim
𝑥→3−
𝑔(𝑥) = lim
𝑥→3+
𝑔( 𝑥) = 𝑔(3)
𝑘√3 + 1 = 𝑚(3) + 2
2𝑘 = 3𝑚 + 2
𝑘 =
3
2
𝑚 + 1 …….(1)
𝑔(𝑥) is differentiable
⇒ 𝐿𝐻𝐷 = 𝑅𝐻𝐷
𝑔′( 𝑥) = {
𝑘
2√ 𝑥 + 1
𝑚
,
,
0 < 𝑥 < 3
3 < 𝑥 < 5
lim
𝑥→3−
𝑔′( 𝑥) = lim
𝑥→3+
𝑔′
(𝑥)
𝑘
2√3 + 1
= 𝑚
𝑘 = 4𝑚
From (1), 4𝑚 =
3
2
𝑚 + 1
5𝑚
2
= 1
⇒ 𝑚 =
2
5
, 𝑘 =
8
5
𝑘 + 𝑚 =
2
5
+
8
5
= 2.
Topic: Limit,Continuity & Differentiability
57. The set of all values of 𝜆 for which the system of linear equations:
2𝑥1 − 2𝑥2 + 𝑥3 = 𝜆𝑥1
2𝑥1 − 3𝑥2 + 2𝑥3 = 𝜆𝑥2
−𝑥1 + 2𝑥2 = 𝜆𝑥3 has a non-trivial solution,

(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer: (B)
Solution:
(2 − 𝜆) 𝑥1 = 2𝑥2 + 𝑥3 = 0
2𝑥1 − ( 𝜆 + 3) 𝑥2 + 2𝑥3 = 0
− 𝑥1 + 2𝑥2 − 𝜆𝑥3 = 0
The systems of linear equations will have a non-trivial solution
⇒ |
2 − 𝜆
2
−1
−2
−𝜆 − 3
2
1
2
−𝜆
| = 0
⇒ (2 − 𝜆)[ 𝜆2
+ 3𝜆 − 4] + 2[−2𝜆 + 2] + 1 4 − 𝜆 − 3] = 0
2𝜆2
+ 6𝜆 − 8 − 𝜆3
− 3𝜆2
+ 4𝜆 − 4𝜆 + 4 + 1 − 𝜆 = 0
−𝜆3
− 𝜆2
+ 5𝜆 − 3 = 0
𝜆3
+ 𝜆2
− 5𝜆 + 3 = 0
( 𝜆 − 1) ( 𝜆2
+ 2𝜆 − 3) = 0
( 𝜆 − 1) ( 𝜆 + 3) ( 𝜆 − 1) = 0
⇒ 𝜆 = 3, −1,−1
Topic: Determinants
58. The normal to the curve, 𝑥2
+ 2𝑥𝑦 − 3𝑦2
= 0, at (1, 1):
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant

(D) Does not meet the curve again
Answer: (C)
Solution:
𝑥2
+ 2𝑥𝑦 − 3𝑦2
= 0
( 𝑥 + 3𝑦) ( 𝑥 − 𝑦) = 0
Pair of straight lines passing through origin.
𝑥 + 3𝑦 = 0 or 𝑥 − 𝑦 = 0
Normal exists at (1, 1), which is on 𝑥 − 𝑦 = 0
⇒ Slope of normal at (1, 1) = −1
∴ Equation of normal will be
𝑦 − 1 = −1 (𝑥 − 1)
𝑦 − 1 = −𝑥 + 1
𝑥 + 𝑦 = 2
Now, find the point of intersection with 𝑥 + 3𝑦 = 0
𝑥 + 𝑦 = 2
𝑥 + 3𝑦 = 0
____________
− 2𝑦 = 2 ⇒ 𝑦 = −1, 𝑥 = 3
(3, 1) lies in fourth quadrant.
Topic: Straight Lines
59. The number of common tangents to the circles 𝑥2
+ 𝑦2
− 4𝑥 − 6𝑦 − 12 = 0 and 𝑥2
+
𝑦2
+ 6𝑥 + 18𝑦 + 26 = 0, is :
(A) 2
(B) 3
(C) 4
(D) 1
Answer: (B)
Solution:

𝑥2
+ 𝑦2
− 4𝑥 − 6𝑦 − 12 = 0
𝐶1(2,3), 𝑟1 = √22 + 32 + 12 = √25 = 5
𝑥2
+ 𝑦2
+ 6𝑥 + 18𝑦 + 26 = 0
𝐶2(−3,−9), 𝑟2 = √32 + 92 − 26
= √90 − 26 = 8
𝐶1 𝐶2 = √52 + 122 = 13
𝐶1 𝐶2 = 𝑟1 + 𝑟2
⇒ Externally touching circles
⇒ 3 common tangents.
Topic: Circle
60. Let 𝑦(𝑥) be the solution of the differential equation( 𝑥 log 𝑥)
𝑑𝑦
𝑑𝑥
+ 𝑦 = 2𝑥 log 𝑥, (𝑥 ≥ 1).
Then 𝑦 (𝑒) is equal to:
(A) 0
(B) 2
(C) 2e
(D) e
Answer: (A)
Solution:
𝑑𝑦
𝑑𝑥
+ (
1
𝑥 log 𝑥
) 𝑦 = 2
Integrating factor = 𝑒∫ 𝑃𝑑𝑥
= 𝑒
∫
1
𝑥 log 𝑥
𝑑𝑥
= 𝑒log(log 𝑥)
= log 𝑥

The solution will be
𝑦 ⋅ 𝑒∫ 𝑃𝑑𝑥
= ∫ 𝑄 ⋅ 𝑒∫ 𝑃𝑑𝑥
+ 𝑐
𝑦 ⋅ log 𝑥 = ∫2 ⋅ log 𝑥 + 𝑐
𝑦 ⋅ log 𝑥 = 2(𝑥 log 𝑥 − 𝑥) + 𝑐
Let 𝑃(1, 𝑦1) be any point on the curve
𝑦1(0) = 2(0 − 1) + 𝑐
𝑐 = 2
When 𝑥 = 𝑒, 𝑦 (log 𝑒) = 2 (𝑒log 𝑒 − 𝑒) + 2
𝑦 = 2
Topic: Differential Equation
Difficulty level: Easy (embibe predicted easy to score (Must Do))

Physics
61. As an electronmakestransitionfromanexcitedstate tothe groundstate of a hydrogenlike
atom/ion:
(A) Kineticenergy,potential energyandtotal energydecrease
(B) Kineticenergydecreases, potential energyincreasesbuttotal energyremainssame
(C) Kineticenergyandtotal energydecreasesbutpotentialenergyincreases
(D) Its kineticenergyincreasesbutpotential energyandtotal energydecrease
Answer:(D)
Solution: 𝑈 = −
𝑒2
4𝜋𝜀0 𝑟
U = potential energy
𝑘 =
𝑒2
8𝜋𝜀0
𝑟
K = kineticenergy
𝐸 = 𝑈 + 𝑘 = −
𝑒2
8𝜋𝜀0 𝑟
𝐸 = Total energy
∴ as electronde-excitesfromexcitedstate togroundstate k increases,U& E decreases
Topic: Modern Physics
Difficulty: Easy(embibe predicted high weightage)
62. The periodof oscillationof asimple pendulumis T = 2π√
L
g
. Measuredvalue of Lis20.0 cm
knownto 1 mm accuracy andtime for 100 oscillationsof the pendulumisfoundtobe 90s
usinga wristwatch of 1s resolution.The accuracyin the determinationof g is:
(A) 3%

(B) 1%
(C) 5%
(D) 2%
Answer: (A)
Solution: ∴ 𝑇 = 2𝜋√
𝐿
𝑔
⇒ 𝑔 = 4𝜋2 𝐿
𝑇2
∴ Error in g can be calculated as
Δ𝑔
𝑔
=
ΔL
𝐿
+
2ΔT
𝑇
∴ Total time for n oscillation is 𝑡 = 𝑛𝑇 where T= time for oscillation.
⇒
Δt
𝑡
=
Δ𝑇
𝑇
⇒
Δ𝑔
𝑔
=
Δ𝐿
𝐿
+
2Δt
𝑡
Given that Δ𝐿 = 1 𝑚𝑚 = 10−3 𝑚, 𝐿 = 20 × 10−2 𝑚
Δ𝑡 = 1𝑠, 𝑡 = 90𝑠
% error in g
Δ𝑔
𝑔
× 100 = (
Δ𝐿
𝐿
+
2Δ𝑡
𝑡
) × 100
=
Δ𝑔
𝑔
× 100 = (
Δ𝐿
𝐿
+
2Δ𝑡
𝑡
) × 100
= (
10−3
20 × 10−2 +
2 × 1
90
) × 100
=
1
2
+
20
9
= 0.5 + 2.22
= 2.72%
≅ 3%

Topic: Unit & Dimensions
Difficulty: Easy (embibe predicted easy to score)
63. A longcylindrical shell carriespositivesurface charge 𝜎 inthe upperhalf andnegative
surface charge – 𝜎 inthe lowerhalf.The electricfieldlinesaroundthe cylinderwill looklike
figure givenin:
(Figuresare schematicandnotdrawn to scale)
(A)
(B)
(C)
(D)

Answer: (D)
Solution:
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle. In the same way we can assume that
negative charge is at centre of mass of that semicircle.
Nowit acts as a dipole nowbythe propertiesof dipole andlowsof electricfieldlinewhere twolinesshould
not intersect the graph would be
Topic: Electrostatics
Difficulty: Moderate (embibe predicted high weightage)

64. A signal of 5 kHzfrequencyisamplitude modulatedonacarrier wave of frequency2MHz.
The frequencyof the resultantsignal is/are:
(A) 2005 kHz, and1995 kHz
(B) 2005 kHz, 2000 kHz and1995 kHz
(C) 2000 kHz and 1995 kHz
(D) 2 MHz only
Answer:(B)
Solution:
Topic: Communication Systems
Difficulty: Easy (embibe predicted high weightage)

65. Consideraspherical shell of radiusRat temperature T.the blackbodyradiationinside itcan
be consideredasan ideal gasof photonswithinternal energyperunitvolume 𝑢 =
𝑈
𝑉
∝
𝑇4and pressure 𝑝 =
1
3
(
𝑈
𝑉
).If the shell now undergoesanadiabaticexpansionthe relation
betweenTandR is:
(A) 𝑇 ∝ 𝑒−3𝑅
(B) 𝑇 ∝
1
𝑅
(C) 𝑇 ∝
1
𝑅3
(D) 𝑇 ∝ 𝑒−𝑅
Answer:(B)
Solution: ∵ in an adiabatic process.
𝑑𝑄 = 0
So by first law of thermodynamics
𝑑𝑄 = 𝑑𝑈 + 𝑑𝑊
⇒ 0 = 𝑑𝑈 + 𝑑𝑊
⇒ 𝑑𝑊 = −𝑑𝑈
∴ 𝑑𝑊 = 𝑃𝑑𝑉
⇒ 𝑃𝑑𝑉 = −𝑑𝑈 ….(i)
Given that
𝑈
𝑉
∝ 𝑇4 ⇒ 𝑈 = 𝑘𝑉𝑇4
⇒ 𝑑𝑈 = 𝑘𝑑( 𝑉𝑇4) = 𝐾( 𝑇4 𝑑𝑉 + 4𝑇3 𝑉 𝑑𝑇)
Also, 𝑃 =
1
3
𝑈
𝑉
=
1
3
𝑘𝑉𝑇4
𝑉
=
𝐾𝑇4
3
Putting these values in equation
⇒
𝐾𝑇4
3
𝑑𝑉 = −𝑘( 𝑇4 𝑑𝑉 + 4𝑇3 𝑉 𝑑𝑇)
⇒
𝑇𝑑𝑉
3
= −𝑇𝑑𝑉 − 4𝑉𝑑𝑇

⇒
4𝑇
3
𝑑𝑉 = −4𝑉𝑑𝑇
⇒
1
3
𝑑𝑉
𝑉
= −
𝑑𝑇
𝑇
⇒
1
3
ln 𝑉 = −ln 𝑇 ⇒ ln 𝑉 = ln 𝑇−3
⇒ 𝑉𝑇3 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
4
3
𝜋𝑅3 𝑇3 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑅𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
⇒ 𝑇 ∝
1
𝑅
Topic: Heat &Thermodynamics
Difficulty: Difficult (embibe predicted high weightage)
66. An inductor(𝐿 = 0.03𝐻) anda resistor(𝑅 = 0.15 𝑘Ω) are connectedinseriestoa battery
of 15V EMF ina circuitshownbelow.The key 𝐾1 hasbeenkeptclosedforalongtime.Then
at 𝑡 = 0. 𝐾1 isopenedandkey 𝐾2 isclosedsimultaneously.At 𝑡 = 1𝑚𝑠,the currentinthe
circuitwill be : ( 𝑒5 ≅ 150)
(A) 67 mA
(B) 6.7 mA
(C) 0.67 mA
(D) 100 mA
Answer: (C)

Topic: Magnetism
67. A pendulummade of auniformwire of crosssectional areaA has time periodT.Whenand
additional massMisaddedto itsbob,the time periodchangesto 𝑇 𝑀.If the Young’s
modulusof the material of the wire isY then
1
𝑌
is equal to:
(g = gravitational acceleration)
(A) [(
𝑇 𝑀
𝑇
)
2
− 1]
𝑀𝑔
𝐴

(B) [1 − (
𝑇 𝑀
𝑇
)
2
]
𝐴
𝑀𝑔
(C) [1 − (
𝑇
𝑇 𝑀
)
2
]
𝐴
𝑀𝑔
(D) [(
𝑇 𝑀
𝑇
)
2
− 1]
𝐴
𝑀𝑔
Answer: (D)
Solution:

Topic: Simple Harmonic Motion
68. A red LED emitslightat 0.1 watt uniformlyaround it.The amplitude of the electricfieldof the lightat a
distance of 1 m from the diode is:

(A) 2.45 V/m
(B) 5.48 V/m
(C) 7.75 V/m
(D) 1.73 V/m
Answer: (A)
Solution:

Topic: Optics
69. Two coaxial solenoidsof differentradii carrycurrentI inthe same direction.Let 𝐹1
⃗⃗⃗⃗ be the
magneticforce onthe innersolenoiddue tothe outerone and 𝐹2
⃗⃗⃗⃗⃗ be the magneticforce on
the outersolenoiddue tothe innerone.Then:
(A) 𝐹1
⃗⃗⃗⃗ is radiallyinwardsand 𝐹2
⃗⃗⃗⃗⃗ isradiallyoutwards
(B) 𝐹1
⃗⃗⃗⃗ is radiallyinwardsand 𝐹2
⃗⃗⃗⃗⃗ = 0
(C) 𝐹1
⃗⃗⃗⃗ is radiallyoutwardsand 𝐹2
⃗⃗⃗⃗⃗ = 0
(D) 𝐹1
⃗⃗⃗⃗ = 𝐹2
⃗⃗⃗⃗⃗ = 0
Answer:(D)
Solution:
𝑆2 is solenoid with more radius than 𝑆1 field because of 𝑆1 on 𝑆2 is o
∴ force on 𝑆2 by 𝑆1 = 0

In the uniform field of 𝑆2 𝑆1 behaves as a magnetic dipole
∴ force on 𝑆1 by 𝑆2 is zero because force on both poles are equal in magnitude and opposite indirection.
Topic: Magnetism
70. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic
expansion,the average timeof collisionbetweenmoleculesincreaseas Vq,where V isthe volumeof the
gas. The value of q is:
(γ =
CP
Cv
)
(A)
3γ−5
6
(B)
γ+1
2
(C)
γ−1
2
(D)
3γ+5
6
Answer: (B)

71. An LCR circuitisequivalenttoadampedpendulum.InanLCR circuitthe capacitor ischarged
to 𝑄0 and thenconnectedtothe L and R as shownbelow:
If a studentplotsgraphsof the square of maximumcharge ( 𝑄 𝑀𝑎𝑥
2 )onthe capacitor withtime (t) fortwo
different values 𝐿1 and 𝐿2(𝐿1 > 𝐿2) of L then which of the following represents this graph correctly?
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer:(D)
Solution:

Topic: Magnetism
72. From a solidsphere of massMand radiusR, a spherical portionof radius
𝑅
2
isremoved,as
showninthe figure.Takinggravitational potential 𝑉 = 0 and 𝑟 = ∞,the potential atthe
centerof the cavitythusformedis:
(G = gravitational constant)
(A)
−𝐺𝑀
𝑅
(B)
−2𝐺𝑀
3𝑅
(C)
−2𝐺𝑀
𝑅
(D)
−𝐺𝑀
2𝑅
Answer:(A)
Solution:

73. A trainis movingona straighttrack withspeed 20 𝑚𝑠−1.It is blowingitswhistle atthe
frequencyof 1000 Hz. The percentage change inthe frequencyheardbya personstanding
nearthe track as the train passeshimis(speedof sound = 320 𝑚𝑠−1) close to:
(A) 12%
(B) 18%
(C) 24%
(D) 6%
Answer:(A)
Solution:

Topic: Wave & Sound
74.
Giveninthe figure are twoblocksA andBof weight20N and100N, respectively.Theseare beingpressed
againsta wall by a force F as shown.If the coefficientof friction betweenthe blocksis0.1 and between
block B and the wall is 0.15, the frictional force applied by the wall on block B is:

(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer: (B)
Solution:
For complete state equilibriumof the system.The state frictionon the blockB by wall will balance the total
weight 120 N of the system.
Topic: Laws of Motion
Difficulty: Moderate (embibe predicted Low Weightage)
75. Distance of the centre of mass of a soliduniformcone itsvertex is z0.If the radiusof itsbase is R and its
height is h then z0 is equal to:
(A)
3ℎ
4
(B)
5ℎ
8
(C)
3𝐻2
8𝑅
(D)
ℎ2
4𝑅
Answer: (A)

Topic: Centre of Mass
Difficulty: Easy (embibe predicted Low Weightage)
76. A rectangularloopof sides10 cm and5cm carrying a current I of 12 A is placedindifferent
orientationsasshowninthe figure below:
If there is a uniformmagneticfieldof 0.3 T inthe positive zdirection,inwhichorientationsthe
loopwouldbe in(i) stable equilibriumand(ii)unstableequilibrium?
(A) (a) and (c),respectively
(B) (b) and (d),respectively
(C) (b) and (c), respectively
(D) (a) and (b),respectively
Answer:(B)
Solution:Fora magneticdipole placedinauniformmagneticfieldthe torque isgivenby 𝜏⃗ =
𝑀⃗⃗⃗ × 𝐵⃗⃗ and potential energyUisgivenas
𝑈 = −𝑀⃗⃗⃗. 𝐵⃗⃗ = −𝑀 𝐵 𝑐𝑜𝑠𝜃
When 𝑀⃗⃗⃗ is inthe same directionas 𝐵⃗⃗ then 𝜏⃗ = 0 and U is min= - MB as 𝜃 = 0 𝑜

⇒ Stable equilibriumis(b).andwhen 𝑀⃗⃗⃗ then 𝜏⃗ = 0 andU ismax = + MB
As 𝜃 = 180 𝑜
Unstable equilibriumin(d)
Topic: Magnetism
Difficulty: Easy(embibe predicted high weightage)
77.
In the circuit shown, the current in the 1Ω resistor is:
(A) 0A
(B) 0.13 A, from Q to P
(C) 0.13 A, from P to Q
(D) 1.3 A, from P to Q
Answer: (B)
Solution:

78. A uniformlychargedsolidsphere of radiusRhaspotential 𝑉0 (measuredwithrespectto ∞)
on itssurface.For thissphere the equipotential surfaceswithpotential
3𝑉0
2
,
5𝑉0
4
,
3𝑉0
4
and
𝑉0
4
have radius 𝑅1, 𝑅2, 𝑅3and 𝑅4 respectively.Then
(A) 𝑅1 ≠ 0 and ( 𝑅2 − 𝑅1) > (𝑅4 − 𝑅3)
(B) 𝑅1 = 0 and 𝑅2 < ( 𝑅4 − 𝑅3)
(C) 2𝑅 < 𝑅4
(D) 𝑅1 = 0 and 𝑅2 > ( 𝑅4 − 𝑅3)
Answer: (B)
Solution:

79. In the given circuit, charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF. Q2 as a
function of ‘C’ is given properly by: (figure are drawn schematically and are not to scale)
(A)
(B)

(C)
(D)
Answer: (A)
Solution:
∵ 1 & 2 𝜇𝑓 are in parallel.
∴ Equivalent capacitance of the series combination is

𝐶 𝑒𝑞 is=
3𝑐
𝐶+3
So total charge supplied by battery is 𝑄 = 𝐶 𝑒𝑞 =
3𝐶𝐸
𝐶+3
∴ Potential difference across parallel combination of 1𝜇𝑓 ad 2𝜇𝑓 is
∆𝑉 =
𝑄
3
=
𝐶𝐸
𝐶+3
So charge on 2𝜇𝑓 capacitor is
𝑄2 = 𝐶2∆𝑉 =
2𝐶𝐸
𝐶+3
⇒
𝑄2
2𝐸
=
𝐶
𝐶+3
⇒
𝑄2
2𝐸
=
𝐶+3−3
𝐶+3
⇒
𝑄2
2𝐸
= 1 −
3
𝐶+3
⇒ (
𝑄2
2𝐸
− 1) = −
3
𝐶+3
⇒ ( 𝑄2 − 2𝐸)( 𝐶 + 3) = −6𝐸
Which is of the form ( 𝑦 − 𝛼)( 𝑥 + 𝛽) < 0
So the graph in hyperbola. With down ward curve line. i.e
80. A particle of massmmovinginthe xdirectionwithspeed 2v ishitbyanotherparticleof mass2mmoving
in the y direction with speed v.If the collision is perfectly inelastic, the percentage loss in the energy
during the collision is close to:
(A) 50%
(B) 56%
(C) 62%
(D) 44%

Answer: (B)
Solution:
The initial momentum of system is 𝑃𝑖
⃗⃗⃗ = 𝑚(2𝑉) 𝑖̂ + (2𝑚) 𝑣 𝑗̂
According to question as
On perfectly inelastic collision the particles stick to each other so.
𝑃𝑓
⃗⃗⃗⃗ = 3𝑚 𝑉𝑓
⃗⃗⃗⃗
By conservation of linear momentum principle
𝑃𝑓
⃗⃗⃗⃗ = 𝑃𝑖
⃗⃗⃗ ⇒ 3𝑚 𝑉𝑓
⃗⃗⃗⃗ = 𝑚2𝑉 𝑖̂ + 2𝑚𝑉𝑗̂
⇒ 𝑉𝑓
⃗⃗⃗⃗ =
2𝑉
3
(𝑖̂ + 𝑗̂) ⇒ 𝑉𝑓 =
2√2
3
𝑉
∴ loss in KE. of system = 𝐾𝑖𝑛𝑖𝑡𝑖 𝑎 𝑙 − 𝐾𝑓𝑖𝑛𝑎𝑙
=
1
2
𝑚(2𝑉)2 +
1
2
(2𝑚) 𝑉2 −
1
2
(3𝑚)(
2√2𝑉
3
)
2
= 2𝑚𝑉2 + 𝑚𝑉2 −
4
3
𝑚𝑉2 = 3𝑚𝑉2 −
4𝑚𝑉2
3
=
5
3
𝑚𝑉2
% change in KE = 100 ×
∆𝐾
𝐾𝑖
=
5
3
𝑚𝑣2
3𝑚𝑉2
=
5
9
× 100
=
500
9
= 56%

Topic: Conservation of Momentum
Difficulty: Moderate (embibe predicted easy to score (Must Do))
81. Monochromaticlightisincidentona glassprismof angle A. If the refractive index of the
material of the prismis 𝜇, a ray,incidentatan angle 𝜃, on the face AB wouldget
transmittedthroughthe face ACof the prismprovided:
(A) 𝜃 < sin−1 [𝜇 sin (𝐴 − sin−1 (
1
𝜇
))]
(B) 𝜃 > cos−1[𝜇 sin (𝐴 + sin−1 (
1
𝜇
))]
(C) 𝜃 < cos−1[𝜇 sin (𝐴 + sin−1 (
1
𝜇
))]
(D) 𝜃 > sin−1 [𝜇 𝑠𝑖𝑛 (𝐴 − sin−1 (
1
𝜇
))]
Answer:(D)
Solution:

82. From a solidsphere of massMand radiusR a cube of maximumpossiblevolumeiscut.
Momentof inertiaof cube aboutan axispassingthroughitscentre and perpendiculartoone
of its facesis:
(A)
𝑀𝑅2
16√2𝜋
(B)
4𝑀𝑅2
9√3𝜋
(C)
4𝑀𝑅2
3√3𝜋
(D)
𝑀𝑅2
32√2𝜋
Answer:(B)
Solution: Let a be length of cube for cube with maximum possible volume diagonal length = 2R
⇒ √3𝑎 = 2𝑅 ⇒ 𝑎 =
2𝑅
√3
As densities of sphere and cube are equal. Let M’ be mass of cube
𝑀
4
3
𝜋𝑅3 =
𝑀′
4
3
𝑎3
𝑀′ =
3𝑀𝑎3
4𝜋𝑅3
Moment of inertial of cube about an axis passing through centre
=
𝑀′(2𝑎2)
12
=
3𝑀𝑎3
4𝜋𝑅3 ×
2𝑎2
12
=
𝑀𝑎5
8𝜋𝑅3

𝑎 =
2
√3
𝑅
=
𝑀 × 32 𝑅5
8𝜋 × 9√3𝑅3
=
4𝑀𝑅2
9√3𝜋
Topic: Rotational Mechanics
Difficulty: Moderate (embibe predicted Low Weightage)
83. Match list-I(FundamentalExperiment) withList-II(itsconclusion) andselectthe correctoptionfromthe
choices given below the list:
List-I List-II
A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron
D (iv) Structure of atom
(A) A − (ii) ;B − (iv);C − (iii)
(B) A − (ii) ;B − (i);C − (iii)
(C) A − (iv) ;B − (iii);C − (ii)
(D)A − (i) ;B − (iv);C − (iii)
Answer: (B)
Solution:Frank-Hertzexperimentdemonstratedthe existence of excitedstatesinmercuryatomshelpingto
confirmthe quantumtheorywhichpredictedthatelectronsoccupiedonlydiscrete quantizedenergystates.
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision.

Davisson-Germer experment = this experiment shows the wave nature of electron.
Topic: Modern Physics
84. When5V potential difference isappliedacrossawire of length0.1 m, the driftspeedof
electronsis 2.5 × 10−4 𝑚𝑠−1.If the electrondensityinthe wire is 8 × 1028 𝑚−3,the
resistivityof the material isclose to:
(A) 1.6 × 10−7Ω𝑚
(B) 1.6 × 10−6 Ω𝑚
(C) 1.6 × 10−5 Ω𝑚
(D) 1.6 × 10−8Ω𝑚
Answer: (C)
Solution:

85. For a simple pendulum, agraphisplottedbetweenitskineticenergy(KE) andpotential
energy(PE) againstitsdisplacementd.whichone of the followingrepresentsthese
correctly?
(Graphs are schematic and not drawn to scale)
(A)

(B)
(C)
(D)
Answer:(A)
Solution:

Topic: Simple Harmonic Motion
86. Two stonesare thrownup simultaneouslyfromthe edge of acliff 240 m highwithinitial
speedof 10 m/sand 40 m/s respectively.Whichof the followinggraphbestrepresentsthe
time variationof relative positionof the secondstone withrespecttothe first?
(Assume stonesdonotreboundafterhittingthe groundandneglectairresistance,take 𝑔 =
10 𝑚𝑠−2)

(the figure are schematicandnot drawnto scale)
(A)
(B)
(C)
(D)
Answer:(B)
Solution: 𝑆1 = 10𝑡 −
1
2
𝑔𝑡2
When 𝑆1 = −240
⇒ −240 = 10𝑡 − 5𝑡2
⇒ 𝑡 = 8𝑠
So at 𝑡 = 8 secondsfirststone will reachground

𝑆2 = 20𝑡 −
1
2
𝑔𝑡2
Till 𝑡 = 8 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑆2 − 𝑆1 = 30𝑡
But after8 second 𝑆1 isconstant = −240
Relative tostone 𝑡1 > 8 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 displacementsof stone 2 𝑆2 + 240
⇒ 𝑆2 + 240 = 20𝑡 −
1
2
𝑔𝑡2
Andat t = 12s secondsstone will reachground
The correspondinggraphof relative positionof secondstone w.r.t.firstis
Topic: Kinematics
Difficulty: Moderate (Embibe predicted high weightage)
87. A solidbodyof constantheatcapacity 1𝐽 ℃⁄ isbeingheatedbykeepingitincontactwith
reservoirsintwoways:
(i) Sequentiallykeepingincontactwith2 reservoirssuchthateachreservoirsuppliessame
amountof heat.
(ii) Sequentiallykeepingincontactwith8 reservoirssuchthateachreservoirsuppliessame
amountof heat
In boththe cases bodyisbroughtfrom initial temperature 100 𝑜 𝐶tofinal temperature 200 𝑜 𝐶.
Entropychange of the bodyin the twocases respectivelyis:
(A) 𝑙𝑛2, 𝑙𝑛2
(B) 𝑙𝑛2,2𝑙𝑛2

(C) 2𝑙𝑛2,8𝑙𝑛2
(D) 𝑙𝑛2,4𝑙𝑛2
Answer: (A)
Solution:

88. Assuminghumanpupil tohave aradiusof 0.25 cm and a comfortable viewingdistance of 25
cm, the minimumseparationbetweentwoobjectsthathumaneye canresolve at500 nm
wavelength is:
(A) 30 𝜇𝑚
(B) 100 𝜇𝑚
(C) 300 𝜇𝑚
(D) 1 𝜇𝑚
Answer:(A)
Solution:
Topic:Optics

89.
Two long current carrying thinwires,bothwith current I, are heldby insulatingthreadsof lengthL and
are in equilibriumasshowninthe figure,withthreadsmakinganangle ‘θ’withthe vertical.If wireshave
mass λ per unit length then the value of I is:
(g = gravitational acceleration)
(A) 2sinθ√
πλgL
μ0 cosθ
(B) 2√
πgL
μ0
tanθ
(C) √
πλgL
μ0
tanθ
(D) 2√
πgL
μ0
tanθ
Answer: (A)
Solution

By equilibrium of mix,
Topic: Magnetism

90. On a hot summernight,the refractive index of airissmallestnearthe groundand increaseswithheight
fromthe ground.Whena lightbeamisdirectedhorizontally,the Huygens’principle leadsustoconclude
that as it travels, the light beam:
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bendsupwards
(D) Becomesnarrower
Answer:(C)
Solution: Consider air layers with increasing refractive index.
At critical angle it will bend upwards at interface. This process continues at each layer, and light ray bends
upwards continuously.
Topic:Optics

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