Measures of Central Tendency: Mean, Median and Mode
Jee main set B all questions
1. 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. 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:
4. (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
Difficulty Level: Easy [Embibe predicted high weightage]
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
5. medicationthatcan control the acid levelsinstomach.Workingof antacids:Antacidscounteract
(neutralize) the acidinstomachthat’susedto aiddigestion.Thishelpsreduce the symptomsof
heartburnandrelievespain.
Topic:Chemistry in everyday life
Difficulty Level: Easy [Embibe predicted easy to score (Must Do)]
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
Difficulty Level: Moderate [Embibe predicted high weightage]
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
6. (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
Difficulty Level: Easy [Embibe predicted high weightage]
10. In the reaction,
The product E is:
(A)
(B)
8. (C) Poly vinyl chloride
(D) Bakelite
Solution: (A) Glyptal is polymer of glycerol and phthalic anhydride.
Topic:Polymers
Difficulty Level: Easy [Embibe predicted easy to score (Must Do)]
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
Difficulty Level: Moderate [Embibe predicted high weightage]
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
9. Solution: (D) Probability of an event involving more than three molecules in a collision are remote.
Topic:Chemical kinetics
Difficulty Level: Easy [Embibe predicted high weightage]
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.
Topic: Group 15 – 18
Difficulty Level: Moderate [Embibe predicted high weightage]
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
Difficulty Level: Easy [Embibe predicted high weightage]
10. 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
Difficulty Level: Moderate [Embibe predicted high weightage]
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
11. Difficulty Level: Easy [Embibe predicted high weightage]
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
Difficulty Level: Easy [Embibe predicted easy to score (Must Do)]
12. 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
Difficulty Level: Easy [Embibe predicted high weightage]
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
13. 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
Difficulty Level: Moderate [Embibe predicted high weightage]
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
Difficulty Level: Easy [Embibe predicted easy to score (Must Do)]
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
14. (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
Difficulty Level: Moderate [Embibe predicted high weightage]
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
Difficulty Level: Easy [Embibe predicted Low Weightage]
25. Which of the following compounds will exhibit geometrical isomerism?
15. (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
Difficulty Level: Easy [Embibe predicted high weightage]
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
16. = 64.38 ≈ 64 g mole⁄
Topic: Liquid solution
Difficulty Level: Easy [Embibe predicted high weightage]
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
Difficulty Level: Easy [Embibe predicted Low Weightage]
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
Difficulty Level: Easy [Embibe predicted Low Weightage]
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
17. (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.
Topic: Group 15 – 18
Difficulty Level: Easy [Embibe predicted high weightage]
18. 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
20. ⇒ 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
21. 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
Difficulty level: Moderate (embibe predicted high weightage)
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:
22. (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.
23. Topic: Parabola
Difficulty level: Moderate (embibe predicted high weightage)
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
Difficulty level: Moderate (embibe predicted high weightage)
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:
25. 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)
26. 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)
34. 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
35. = 𝑙𝑛 × (2𝑚)2 [∵ 𝑓𝑟𝑜𝑚 (1)]
= 4𝑙𝑚2
𝑛
Topic: Progression & Series
Difficulty level: Moderate (embibe predicted high weightage)
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
36. 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:
37. 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
Difficulty level: Easy (embibe predicted high weightage)
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
38. (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
Difficulty level: Moderate (embibe predicted high weightage)
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
39. 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)
43. (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
Difficulty level: Easy (embibe predicted high weightage)
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)
50. 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))
51. 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%
52. (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%
54. 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)
55. 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)
56. 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
57. ⇒
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)
59. Topic: Magnetism
Difficulty: Moderate (embibe predicted high weightage)
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
61. Topic: Simple Harmonic Motion
Difficulty: Difficult (embibe predicted high weightage)
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:
63. Topic: Optics
Difficulty: Easy (embibe predicted high weightage)
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
64. 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
Difficulty: Moderate (embibe predicted high weightage)
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)
66. 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:
67.
68. Topic: Magnetism
Difficulty: Difficult (embibe predicted high weightage)
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:
70. Difficulty: Moderate (embibe predicted high weightage)
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:
71. Topic: Wave & Sound
Difficulty: Moderate (embibe predicted high weightage)
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:
72. (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)
74. 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
75. ⇒ 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:
77. 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:
78.
79. Topic: Electrostatics
Difficulty: Moderate (embibe predicted high weightage)
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)
81. 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
Topic: Electrostatics
Difficulty: Moderate (embibe predicted high weightage)
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%
82. 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%
83. 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:
85. Difficulty: Moderate (embibe predicted high weightage)
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
86. 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.
87. Davisson-Germer experment = this experiment shows the wave nature of electron.
Topic: Modern Physics
Difficulty: Easy (embibe predicted high weightage)
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:
88. Topic: Electrostatics
Difficulty: Easy (embibe predicted high weightage)
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)
94. 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
95. Difficulty: Moderate (embibe predicted high weightage)
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
97. Difficulty: Moderate (embibe predicted high weightage)
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
Difficulty: Moderate (embibe predicted high weightage)