datavorous/entrance-exam-dataset huggingface dataset:
$\int \frac{d x}{\cos x-\sin x}$ is equal to

Options:
A
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{8}\right)\right|+C$

B
$\frac{1}{\sqrt{2}} \log \left|\cot \left(\frac{x}{2}\right)\right|+C$

C
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{3 \pi}{8}\right)\right|+C$

D
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$

The correct answer is:
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$

Solution:
$\int \frac{d x}{\cos x-\sin x}=\frac{1}{\sqrt{2}} \int \frac{1}{\cos \left(x+\frac{\pi}{4}\right)} d x=\frac{1}{\sqrt{2}} \int \sec \left(x+\frac{\pi}{4}\right) d x=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2004
Cp-Cv=RM and Cv are specific heats at constant pressure and constant volume respectively. It is observed that, Cp-Cv=a for hydrogen gas and Cp-Cv=b for nitrogen gas. The correct relation between a and b is:

Options:
A a=28 b
B a=114 b
C a=b
D a=14b

The correct answer is:
a=14b

Solution:
Cp-Cv=RM
For hydrogen Cp-Cv=R2=a
For nitrogen Cp-Cv=R28=b
ab=14
a=14b

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2017 (02 Apr)
Cr2+ and Mn3+ both have d4 configuration. Thus

Options:
A both are reducing agents
B both are oxidizing agents
C Cr2+ is an oxidizing agent while Mn3+ is a reducing agent.
D Mn3+ is an oxidizing agent while Cr2+ is a reducing agent.

The correct answer is:
Mn3+ is an oxidizing agent while Cr2+ is a reducing agent.

Solution:
Mn3++e-→Mn2+
Extra stability is gained when Mn3+ is reduced to Mn2+ and is thus an oxidizing agent.
Cr2+→Cr3++e-

Topic: Chemistry, d and f Block Elements, JEE Main
CrO3 dissolves in aqueous NaOH to give

Options:
A CrO42-
B CrOH3-
C CrO72-
D CrOH2

The correct answer is:
CrO42-

Solution:
CrO3 dissolves in aqueous NaOH to give sodium chromate.
CrO3+2NaOH â¶Na2CrO4+H2O
Sodium chromate

Topic: Chemistry, d and f Block Elements, JEE Main
$C s B r$ crystal has bcc structure. It has an edge length of 4.3 Å. The shortest interionic distance between $\mathrm{Cs}^{+}$and $\mathrm{Br}^{-}$ions is

Options:
A $1.86 Å$
B $3.72 Å$
C $4.3 Å$
D $7.44 Å$

The correct answer is:
$3.72 Å$

Solution:
Closest approach in bcc lattice
$=\frac{1}{2}$ of body diagonal $=\frac{1}{2} \times \sqrt{3} a=\frac{\sqrt{3}}{2} \times 4.3=3.72 Å$

Topic: Chemistry, Solid State, JEE Main
C → strongly heated B ← 200 o C CaSO 4 ·2H 2 O → Heat 120 o C A
A, B and C are respectively

Options:
A plaster of paris, dead burnt plaster, calcium sulphide
B dead burnt plaster, plaster of Paris, lime
C plaster of Pairs, dead burnt plaster, calcium sulphite
D plaster of Paris, dead burnt plaster, calcium oxide (lime)

The correct answer is:
plaster of Paris, dead burnt plaster, calcium oxide (lime)

Solution:

Topic: Chemistry, s Block Elements, JEE Main
Ca has atomic no. 20 and atomic weight 40 . Which of the following statements is not rrect about atom

Options:
A The number of electrons is same as the number of neutrons
B The number of nucleons is double of the number of electrons
C The number of protons is half of the number of neutrons
D The number of nucleons is double of the atomic number

The correct answer is:
The number of protons is half of the number of neutrons

Solution:
Calcium has has 20 proton, 20 neutron

Topic: Chemistry, Structure of Atom, JEE Main
Caffeine has a molecular weight of 194 . If it contains $28.9 \%$ by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is

Options:
A $4$
B $6$
C $2$
D $3$

The correct answer is:
$4$

Solution:
$100 \mathrm{gm}$ caffeine has $28.9 \mathrm{gm}$ nitrogen
$\begin{aligned} 194 \mathrm{gm} \text { caffeine has }=\frac{28.9}{100} \times 194 & =56.06 \mathrm{gm} \\ \therefore \quad \text { No. of atoms in } \quad & =\frac{56.06}{14} \approx 4 \text { caffeine }\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calamine and Malachite, respectively, are the ores of :

Options:
A Copper and Iron
B Zinc and Copper
C Aluminium and Zinc
D Nickel and Aluminium

The correct answer is:
Zinc and Copper

Solution:
Malachite ⇒CuCO3·Cu(OH)2

Calamine ⇒ZnCO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Calamine is an ore of:

Options:
A Iron
B Copper
C Aluminium
D Zinc

The correct answer is:
Zinc

Solution:
Calamine is zinc carbonate ZnCO3. It is ore of zinc.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2015 (11 Apr Online)
Calcination is the process in which:

Options:
A
ore is heated above its melting point to expel $\mathrm{H}_2 \mathrm{O}$ or $\mathrm{CO}_2$ or $\mathrm{SO}_2$

B
ore is heated below its melting point to expel volatile impurities

C
ore is heated above its melting point to remove $\mathrm{S}, \mathrm{As}$ and $\mathrm{Sb}$ as $\mathrm{SO}_2, \mathrm{As}_2 \mathrm{O}_3$ and $\mathrm{Sb}_2 \mathrm{O}_3$ respectively

D
ore is heated below its melting point to expel $\mathrm{H}_2 \mathrm{O}$ or $\mathrm{CO}_2$

The correct answer is:
ore is heated below its melting point to expel $\mathrm{H}_2 \mathrm{O}$ or $\mathrm{CO}_2$

Solution:
Calcination is a process of heating a substance to a high temperature but below the melting or fusion point, causing loss of moisture, reduction or oxidation and dissociation into simpler substances.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2013 (09 Apr Online)
Calcitonin

Options:
A is secreted by the parathyroid glands
B causes blood calcium levels to decrease
C insufficiency results in weak bones and tetany
D levels increase when blood calcium level decrease

The correct answer is:
causes blood calcium levels to decrease

Solution:
The thyroid is the largest endocrine in our body. It is butterfly-shaped and situated in the neck region. It consists of two lobes on either side of trachea that are connected by isthmus. Each lobe consists of many follicles that secrete thyroid hormones that control the metabolism of the body. In between follicles are parafollicular cells that secrete calcitonin. On the posterior side of thyroid, two pairs of parathyroid glands are seen and these secrete PTH (parathormone). Calcitonin and parathormone maintain calcium metabolism. Calcitonin causes blood calcium level to decrease. This is done by deposition of calcium in bones, increasing absorption of calcium by kidneys etc.

Topic: Biology, Chemical Coordination and Integration, NEET
Calcium butanoate on heating followed by tratment with 1,2 -ethanediol in the presence of catalytic amount of acid, produces a major product which is

Options:
A
B
C
D

The correct answer is:

Solution:
No Solution

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Calcium carbide on reacting with water gives

Options:
A Methane
B Ethane
C Ethene
D Acetylene

The correct answer is:
Acetylene

Solution:
$\mathrm{CaC}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { Actylene }}{\mathrm{CH}} \equiv \mathrm{Ca}(\mathrm{OH})_2$

Topic: Chemistry, Hydrocarbons, JEE Main
Calcium crystallizes in a face centred cubic unit cell with a =â0.560ânm. The density of the metal if it contains 0.1% schottky defects would be:

Options:
A 1 .51 gcm -3
B 2 .51 gcm -3
C 15 .1 gcm -3
D 0 .151 gcm -3

The correct answer is:
1 .51 gcm -3

Solution:
A face centred cubic unit cell contains four atoms. Schottky defect is basically a vacancy defect in ionic solids. Due to Schottky defects, the vacant spaces will increase, resulting in decrease in number of atoms per unit cell.
Thus, number of atoms will reduce to:
Z= 4â 4Ã0.1 100 =3.996
The density of the metal can be calculated by:
Ïâ=â ZâÃâM N 0 âÃâ a 3
Where, M is the molecular mass of calcium = 40 g mol -1
N 0 is the Avogadro number =6.022Ã 10 23 mol - 1
a is the edge length = 0.560 nm â= â0.56âÃâ 10 â7 cm
Putting the values in the formula above:
=â 3.996âÃâ(40âgâ mol â1 ) (0.560âÃâ 10 â7 âcm) 3 âÃâ(6.022âÃâ 10 23 â mol â1 )
=1.51âgâ cm â3

Topic: Chemistry, Solid State, JEE Main
Calcium imide on hydrolysis gives gas B which on oxidation by bleaching powder gives gas C. Gas C on reaction with magnesium give compound D which on hydrolysis gives again gas B. Identify B, C and D respectively.

Options:
A NH3,N2,Mg3N2
B N2,NH3,MgNH
C N2,N2O5,Mg(NO3)2
D NH3,NO2,Mg(NO2)2

The correct answer is:
NH3,N2,Mg3N2

Solution:
Here B, C and D are NH3,N2 and Mg3N2 respectively
CaNH+2H2O→CaOH2+NH3(g)(B)
2NH3(B)+3CaOCl2→N2(g)(C)+3CaCl2+3H2O
N2(g)(C)+3Mg→Mg3N2(D)
Mg3N2(D)+6H2O→3MgOH2+2NH3(B)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Calcium lactate is a salt of weak acid i.e., lactic acid having general formula Ca(LaC)2 . Aqueous solution of salt has 0.3 M concentration. pOH of solution is 5.60. If 90% of the salt is dissociated then what will be the value of pKa ?

Options:
A 2.8 – log (0.54)
B 2.8 + log (0.54)
C 2.8 + log (0.27)
D None of these

The correct answer is:
2.8 – log (0.54)

Solution:
Ca(LaC)2âCa2++2LaC-
Concentration of salt =0.62=0.3 M
Concentration of LaC- after dissociation =2Ã0.3Ã0.9=0.54 M
pH after salt hydrolysis may be calculated as,
pH =  1 2 [(pK w  +  pK a  +  log(LaC − )]
14-5.6=12[14+pKa+logâ¡0.54]
pKa=2.8-logâ¡0.54

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calcium plays a very important role in the formation of bones. Write on the role of endocrine glands and hormones responsible for maintaining calcium homeostasis.

Options:

None

Solution:
The hormones and endocrine glands that are responsible for maintaining calcium homeostasis, are thyroid and parathyroid glands and their associated hormones are calcitonin and Parathyroid Hormone (PTH).
(i) Parathyroid glands - These glands developed from the endoderm of the embryo. The cells of parathyroid glands are of two types - chief cells and oxyphil cells. The chief cells of the parathyroid glands secrete parathyroid hormone $(\mathrm{PTH})$.
This hormone (PTH) is involved in regulation of calcium and phosphate balance between the blood and other tissue. It mobilises the release of calcium into the blood from bones. PTH increases reabsorption of calcium by the body organs like intestine and kidneys.
(ii) Thyroid gland - It is the largest endocrine gland located anterior to the thyroid cartilage of the larynx in the neck. This gland plays a major role in maintaining calcium homeostasis. It releases thyrocalcitonin hormone produced by the parafollicular cells, also called, ' $\mathrm{C}$ ' cells.
This hormone is secreted when the calcium level in blood gets high. It is a 32 amino acid peptide hormone that lowers the calcium level bysuppressing release ofcalcium ions from the bones. Calcitonin thus has an action opposite to that of the parathyroid hormone in calcium homeostasis.

Topic: Biology, Chemical Coordination and Integration
Calculate dissociation constant for an acid HA having [H+] equals to 0.1 M (Given: degree of dissociation of HA is 1% )

Options:
A 10-8
B 10-6
C 10-3
D 10-7

The correct answer is:
10-3

Solution:
ka=Î±2C1-Î± , Î± is very small ;
â´1-Î±â1
CÎ±â=â0.1
k a â=â(CÎ±)Î±
=0.01×0.1= 10 −3

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate ÎGo for the following cell reaction
Zn s + Ag 2 O s + H 2 O l â Zn 2+ aq +2Ag s +2 OH â aq

EAg+/Ag0=+0.80V and EZn+2/Zn0=-0.76V ,F=96500

(Given: K sp ofâAgOHâ=â2Ãâ 10 â8 )

Options:
A -305kJ/mol
B 212 kJ/mol
C 305Â kJ/mol
D 301Â kJ/mol

The correct answer is:
212 kJ/mol

Solution:
Calculation for EOH-Ag2OAg-
Reduction [2Ag++2e-â2Ags,âG1o=-2ÃFÃEoAg+|Ag
Ionisation [Ag2O+H2Olâ2Ag++2OH-âG2o=-2.303RTlogksp2
Net Ag2O+H2Ol+2e-â 2Ags+2OH-,âG3o=-2ÃFÃEOH-Ag2OAgo
âG3o=âG1o+âG2o
+2ÃFÃEOH-Ag2OAgo=+2ÃFÃEoAg+|Ag+2.303RTlogksp22F
EOH-Ag2OAgo=EoAg+|Ag+0.062Ã2logâ¡ksp
=0.8+0.06logâ¡2Ã10-8
=0.8+0.06-8+0.3
=0.8+0.06Ã-7.7
=0.8-0.462
=0.338Voltâ0.34V âGo=-nFEceuo
=-nFEco-EARed
=-2ÃEOH-Ag2OAgo-EZn+2|Zno
=-2Ã96500Ã0.34+0.76
=-2Ã96500Ã1.1volt
=212300J
=212.3 kJ

Topic: Chemistry, Electrochemistry, JEE Main
Calculate H+ and % dissociation of 0.1 M solution of NH4OH solution. The ionisation constant for NH4OH is kb=2.0×10-5

Options:
A 7.09×10-12M,3%
B 7.09×10-12M,1.4%
C 9.02×10-12M,2.4%
D 9.02×10-12M,3%

The correct answer is:
7.09×10-12M,1.4%

Solution:
kb=(Cα)2C1-α=Cα21-α
negligible 1-α=1
kb=Cα2
α=kbC=2×10-50.1=2×10-4
α=1.41×10-2
α=1.4%
OH-=Cα=0.1×1.4×10-2
=1.4×10-3M
H-=KwOH-=10-141.4×10-3
=7.09×10-12M

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate ÎHfo for chloride ion from the following data:

1 2 H 2 g + 1 2 Cl 2 g âHCl g ; ÎH f o =â92.4 kJ

HCl g + nH 2 Oâ H + aq + Cl â aq ; ÎH f o = â74.8 kJ

ÎÎ f o ââofââ H + â aq =0.0ââkJ

Options:
A -167.2 kJ
B -165.2Â kJ
C -157.2Â kJ
D -147.2Â kJ

The correct answer is:
-167.2 kJ

Solution:
Given,
1 2 H 2 g + aq â H + aq + e â ; ÎH o =0â¦. i
1 2 H 2 g + 1 2 Cl 2 g âHCl g ; ÎH o = â92.4 kJâ¦ ii
HCl g + nH 2 O l → H + aq + Cl − aq ; ΔH o = −74.8 kJ… iii HCl g + nH 2 O l → H + aq + Cl − aq ; ΔH o = −74.8 kJ… iii
ii ââ+ââ iii âââââ i
â´ 1 2 Cl 2 g + aq + e â â Cl â aq ;ÎH= â92.4â74.8 kJâ167.2 kJâ¦. iv
Heat of formation for Cl â aq =â167.2 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate $\wedge_{H O A c}^{\infty}$ Using appropriate molar conductances of the electrolytes listed above at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ at $25^{\circ} \mathrm{C}$

Options:
A
517.2

B
552.7

C
390.7

D
217.5

The correct answer is:
390.7

Solution:
$\wedge_{\mathrm{AcOH}}^{\infty}=\wedge_{\mathrm{HCl}}^{\infty}+\wedge_{\mathrm{AcONa}}^{\infty}-\wedge_{\mathrm{NaCl}}^{\infty}$
$=390.7$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2005
Calculate heat of atomization of furan in KJmol-1 using the data

Heats of atomization of C, H, O are 717, 218, 249 KJmol-1 each. Fill your answer dividing by 100 If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
81

Solution:
Writing the formation reaction
4C s +4H g +O g â C 4 H 4 O g
Î H f â = - 6 2 Â· 0 = Î£ reac BE - Î£ product BE
Î£ product BE = Î£ reac BE + 6 2 Â· 0
= 4 Ã 7 1 7 + 4 Ã 2 1 8 + 2 4 9 + 6 2
= 2 8 6 8 + 8 7 2 + 2 4 9 + 6 2 = 4 0 5 1 KJ mol - 1
Î£ product BE is heat of atomization of product into isolated gaseous atoms.
Hence answer is 40.51

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate heat of atomization of furan in KJmol-1 using the data

Heats of atomization of C, H, O are 717, 218, 249 KJmol-1 each. Fill your answer dividing by 100

Options:

The correct answer is:
40.51

Solution:
Writing the formation reaction
4C s +4H g +O g â C 4 H 4 O g
Î H f â = - 6 2 Â· 0 = Î£ reac BE - Î£ product BE
Î£ product BE = Î£ reac BE + 6 2 Â· 0
= 4 Ã 7 1 7 + 4 Ã 2 1 8 + 2 4 9 + 6 2
= 2 8 6 8 + 8 7 2 + 2 4 9 + 6 2 = 4 0 5 1 KJ mol - 1
Î£ product BE is heat of atomization of product into isolated gaseous atoms.
Hence answer is 40.51

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate $\Lambda_{\text {HOAC }}^{\infty}$ using appropriate molar conductances of the electrolytes listed above at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ at $25^{\circ} \mathrm{C}$

Options:
A $517.2$
B $552.7$
C $390.7$
D $217.5$

The correct answer is:
$390.7$

Solution:
$\Lambda_{\mathrm{HOAC}}^{\infty}=\Lambda_{\mathrm{NeOAC}}^{\infty}+\Lambda_{\mathrm{HCl}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty}=91.0+426.2-126.5=390.7$

Topic: Chemistry, Electrochemistry, JEE Main
Calculate how much population would be covered by T.V. broadcast if height of TV tower is 100 m and the average population density around the tower is 1000 km-2 (Take radius of earth as R=6.4Ã 10 6 m )

Options:
A 6Ã106
B 2×106
C 12×106
D 4×106

The correct answer is:
4×106

Solution:
Radius of range covered by tower, d=2hR

Population covered =ρ×πd2=ρ×π(2hR)=4×106

Topic: Physics, Communication System, JEE Main
Calculate $\mathrm{K}_{\mathrm{C}}$ for the reversible process given below if $\mathrm{K}_{\mathrm{P}}=167$ and $\mathrm{T}=800^{\circ} \mathrm{C}$
$\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$

Options:
A 1.95
B 1.85
C 1.89
D 1.60

The correct answer is:
1.89

Solution:
Hints: $K_p=K_C(R T)^{\Delta n}$
for $\mathrm{eq}^{\mathrm{n}} \quad \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{n}=1$
$\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{167}{(0.0821 \times 1073)^1}=1.89$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Calculate \( \left[\mathrm{OH}^{-}\right] \times 10^{8} \) for a solution whose \( \mathrm{pH} \) is \( 6.2 \).
GIve your answer up to two significant figures.

Options:

The correct answer is:
1.6

Solution:
Given: pH=6.2

In the pH scale,

pH+pOH=14pOH=14-6.2=7.8pOH=-logOH-=7.8⇒logOH-=-7.8=0.2-8OH-=1.6×10-8

1.6×10-8M

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate milli equivalent of washing soda required to remove its hardness from one litre hard water contains 18.00 mg Mg2+

Options:

The correct answer is:
1.5

Solution:
Meq of Mg+2≡ Meq of washing soda
WE×1000=Mg+2 ; EW=242=12
18×10-312×1000=1.5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate normality, molarity and molality of the solution containing 22% (w/w) salt Al2(SO4)3 d=1.253 gm/ml by weight is

Options:
A 4.83 N, 8.25 M, 8.25 m
B 48.3Â N,Â 0.825Â M,Â 0.825Â m
C 4.83Â N,Â 4.83Â M,Â 4.83Â m
D 4.83 N, 0.805 M,0.825 m

The correct answer is:
4.83 N, 0.805 M,0.825 m

Solution:
Normality =w/w %Ã10ÃdE2 =22Ã10Ã1.2533426=4.83 N
Molarity =w/w %Ã10ÃdM2=22Ã10Ã1.253342=0.805 M

E2= eq.mass of solute

M2= molar ass of solute

Molality =w/w%Ã1000M2100-w/w%
Molality =22Ã1000342(100-22)=0.825 m

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate pH at which an acid indicator HIn with concentration 0.1 M changes its colour ( Ka for HIn =1×10-5 ).

Options:

The correct answer is:
5

Solution:
Indicator equation is
pH=pKIn+log⁡In-HIn
Colour change takes place when [In-]=[HIn]
i.e., when pH=pKIn=5 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate Δ r G for the reaction at 27 o C

H 2 g + 2 Ag + aq ⇌ 2 Ag s + 2 H + aq

Given : P H 2 = 0.5 bar ; Ag + = 10 - 5 M ;

H + = 1 0 - 3 M ; Δ f G ∘ Ag + aq = 77.1 kJ/mol

Options:
A - 154.2 kJ/mol
B - 178.9 kJ/mol
C - 129.5 kJ/mol
D None of these

The correct answer is:
- 129.5 kJ/mol

Solution:
Δ r G â = 0 - 77.1 × 2 = - 154.2 kJ/mol
Q = H + 2 P H 2 · Ag + 2
= 1 0 - 6 0.5 × 1 0 - 1 0 = 2 × 1 0 4
Δ G = Δ r G â + RT ln Q
Δ r G = - 154.2 + 8.314 × 300 ln 2 × 1 0 4 1 0 0 0
=- 129.5 kJ/mol

Topic: Chemistry, Chemical Equilibrium, JEE Main
Calculate standard free energy change for the reaction $\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}^{2+}+\mathrm{Cl}^{-}$taking place at $25^{\circ} \mathrm{C}$ in a cell whose standard e.m.f. is 1.02 volts

Options:
A $-98430 J$
B $98430 \mathrm{~J}$
C $96500 \mathrm{~J}$
D $-49215 J$

The correct answer is:
$-98430 J$

Solution:
$\Delta G=-n F E^O$
$\Delta G=-1 \times 96500 \times 1.02 . \quad \Delta G=-98430$

Topic: Chemistry, Electrochemistry, JEE Main
Calculate standard free energy change for the reaction 2Ag+2H+→H2+2Ag+

Given: EAg+/Ago=+0.80V

Options:
A 308.08 kJ
B 154.4 kJ
C 77.2 kJ
D – 154.4 kJ

The correct answer is:
154.4 kJ

Solution:
Ecello=EH2o-EAgo
=0-0.80=-0.80
ÎGÂ°=-nFEcello
=-2Ã96500Ã(-0.8)
=+154.4 kJ

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly and the surface is 0.05. (g=10 m s-2, mass of the string is negligible and no other friction exists).

Options:
A 1.25 $ms^{-2}$
B 1.50 $ms^{-2}$
C 1.66 $ms^{-2}$
D 1.00 $ms^{-2}$

The correct answer is:
1.25 $ms^{-2}$

Solution:
First, make the free body diagram,

here, the equation, the relation between tension and acceleration, 20=2a+T ..1 & T=10a+5 ..2⇒20=2a+10a+5⇒12a=15⇒a=1512=1.25 m s-2

Topic: Physics, Laws of Motion, JEE Main
Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K.

Options:
A 3.45 kJ/mol
B 6.90 kJ/mol
C 9.68 kJ/mol
D 19.6 kJ/mol

The correct answer is:
3.45 kJ/mol

Solution:
Relation between temperature and activation energy is given by Arrhenius equation.

k =Ae-EaRT

k2k1 =e-EaRT2+EaRT1ln k2k1=-EaRT2+EaRT1ln 2 =EaR1T1-1T2log 2=Ea2.303R1300-16000.3010=Ea2.303×8.3141600Ea=3.45 kJ/mol

Topic: Chemistry, Chemical Kinetics, JEE Main
Calculate the amount of bromine required to convert 10 g of phenol into 2, 4, 6-tribromophenol (Report your answer as whole number only)

Options:

The correct answer is:
51

Solution:
PhOH +â3Br2âC6H2Br3OH+3HBr
1 mole of phenol require 3 moles of Br 2
94 g of phenol reacts with 480âg of Br 2 .
10âgâofâphenolâââ 480 94 âÃâ10â=â51âg

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P4O10.

Options:

The correct answer is:
1008

Solution:
6 CaO + P 4 O 1 0 ⟶ 2 Ca 3 PO 4 2
1 Mole P4O10 = 284 g/mole
Moles of P4O10 = 8 5 2 2 8 4 = 3 moles
Moles of CaO = 18 moles
Mass of CaO = 18 x 56 = 1008 g

Topic: Chemistry, Redox Reactions, JEE Main
Calculate the amount of heat (in calories) required to convert $5 \mathrm{gm}$ of ice at $0^{\circ} \mathrm{C}$ to steam at $100^{\circ} \mathrm{C}$

Options:
A $3100$
B $3200$
C $3600$
D $4200$

The correct answer is:
$3600$

Solution:
Ice $\left(0^{\circ} \mathrm{C}\right)$ converts into steam $\left(100^{\circ} \mathrm{C}\right)$ in following three steps.
Total heat required $Q=Q_1+Q_2+Q_3$
$=5 \times 80+5 \times 1 \times(100-0)+5 \times 540=3600 \mathrm{cal}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. $\left(\mathrm{k}=0.02303\right.$ minute $\left.^{-1}\right)$

Options:
A $25 \%$
B $50 \%$
C $75 \%$
D $12.5 \%$

The correct answer is:
$25 \%$

Solution:
For a first order reaction,
$\begin{aligned}
& k=\frac{0.693}{t_{1 / 2}} \\
& t_{1 / 2}=\frac{0.693}{0.02303}=30 \mathrm{~min}
\end{aligned}$
Percent of reactant that remains after $t_{1 / 2}=50 \%$. $2 \times \mathrm{t}_{1 / 2}=60 \mathrm{~min}$
Therefore, percent of reactant that remains after $2 \mathrm{t}_{1 / 2}=25 \%$

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2023 (09 May Shift 1)
Calculate the angular frequency of the system shown in figure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that mA = mB = m.

Options:
A 2k 4m
B 4k 5m
C 6k 7m
D 8k 5m

The correct answer is:
4k 5m

Solution:
Let x0 be the extension in the spring in equilibrium. Then equilibrium of A and B give,
T = kx0 + mg sin θ ...(i)
and 2T = mg ...(ii)
Here, T is the tension in the string. Now, suppose A is further displaced by a distance x from its mean position and v be its speed at this moment. Then B lowers by x 2 and speed of B at this instant will be v 2 Total energy of the system in this position will be,
E = 1 2 k x + x 0 2 + 1 2 m A v 2 + 1 2 m B v 2 2 + m A gh A - m B gh B
or E = 1 2 k x + x 0 2 + 1 2 m v 2 + 1 8 mv 2 + mgx sin θ - mg x 2
or E = 1 2 k x + x 0 2 + 5 8 m v 2 + mgx sin θ - mg x 2
Since, E is constant,
dE dt = 0
or 0 = k x + x 0 d x dt + 5 4 mv dv dt + mg sin θ d x dt - mg 2 d x dt
Substituting, d x dt = v
d v dt = a
and k x 2 + mg sin θ = mg 2 From equations (i) and (ii)
We get, 5 4 m a = - k x
Since, a ∝ - x
Motion is simple harmonic, time period of which is,
T = 2 π | x a | = 2 π 5m 4k
ω = 2 π T = 4k 5m

Topic: Physics, Oscillations, JEE Main