datavorous/entrance-exam-dataset huggingface dataset:
Consider the figure provided.
\(1 \mathrm{~mol}\) of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at \(18^{\circ} \mathrm{C}\). If the piston is moved to position \(\mathrm{B}\), keeping the temperature unchanged, then ' \(x\) ' \(L\) atm work is done in this reversible process.
\(x=\) ______ \(\mathrm{L}\) atm. (nearest integer)
[Given : Absolute temperature \(={ }^{\circ} \mathrm{C}+273.15, \mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}\) ]

Options:

The correct answer is:
55

Solution:
$\begin{aligned} & \omega=-n R T \ln \left(\frac{V_2}{V_1}\right) \\ & =-1 \times \cdot 08206 \times 291 \cdot 15 \ln \left(\frac{100}{10}\right) \\ & =-55.0128\end{aligned}$
Work done by system $\approx 55 \mathrm{~atm}$ lit.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (08 Apr Shift 1)
Consider the following acids
1.MeCH2COOH
2.Me2CHCOOH
3.Me3CCOOH
4.Et3CCOOH
Correct order of the rate of esterification of these acids with MeOH is

Options:
A 1 > 2 > 3 > 4
B 2 > 1 > 3 > 4
C 2 > 3 > 4 > 1
D 2 > 3 > 1 > 4

The correct answer is:
1 > 2 > 3 > 4

Solution:
The correct order of the rate of esterification of these acids with MeOH is 1 > 2 > 3 > 4.
The mechanism of esterification is

As the size of the substitution on α -carbon increases, the tetrahedral bonded intermediate becomes more crowded. The greater the crowding, the slower is the reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the following antibiotics
(i) Erythromycin
(ii) Ofloxacin
(iii) Chloramphenicol
(iv) Penicillin
The pair of bactericidal antibodies is

Options:
A i-iii
B ii-iv
C ii-v
D i-iv

The correct answer is:
ii-iv

Solution:
Antibiotics are the substances which in low concentration inhibit the growth or destroy microorganisms by intervening in their metabolic processes. They are called bactericidal when they kill the microbes, e.g., penicillin, aminoglycosides, ofloxacin, etc.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Consider the following bromides :
The correct order of $\mathrm{S}_{\mathrm{N}} 1$ reactivity is

Options:
A
$B>C>A$

B
$B>A>C$

C
$C>B>A$

D
$A>B>C$

The correct answer is:
$B>C>A$

Solution:
$\mathrm{S}_{\mathrm{N}} 1$ proceeds via carbocation intermediate, the most stable one forming the product faster. Hence reactivity order for $\mathrm{A}, \mathrm{B}, \mathrm{C}$ depends on stability of carbocation created.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2010
Consider the following bromides

The correct order of SN1 reactivity is

Options:
A A > B > C
B B > C > A
C B > A > C
D C > B > A

The correct answer is:
B > C > A

Solution:
SN1 reaction rate depends upon the stability of the carbocation, as carbocation formation is the rate determining step compound (ii), forms at 2° allylic carbocation which is the most stable, the next stable carbocation is formed from (iii), it is a 2° carbocation, (i) forms the least stable 1° carbocation, the order of reactivity is thus,

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Consider the following carbanions

1. CH3-C⊖H2 2. CH2=C⊖H 3. CH≡C⊖

Correct the order of stability of these carbanions in decreasing order is

Options:
A 1 > 2 > 3
B 2 > 1 > 3
C 3 > 2 > 1
D 3 > 1 > 2

The correct answer is:
3 > 2 > 1

Solution:
Stability of carbanion (when hybridisation of C⊖ is different) is directly proportional to S character.
Stability ∝ s%

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following carbanions:

Correct decreasing order of stability is -

Options:
A II > III > IV > I
B III > IV > I > II
C IV > I > II > III
D I > II > III > IV

The correct answer is:
II > III > IV > I

Solution:
carbanion stability ââMââIâ 1 +I/+HC â 1 +M
In (II), stability is increased as there is an electron withdrawing group, hence it stabilizes, but in (I) and (IV), it increases the electron density whereas in (III), there is no functional group attached for electron donation or withdrawal.

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following carbocations, stability of carbocations increasing order
I C 6 H 5 C + H 2 ââ II â C 6 H 5 CH 2 C + H 2
III â C 6 H 5 C + HCH 3 â IV â C 6 H 5 C + CH 3 2

Options:
A (II) <(I) <(III) <(IV)
B (I) <(II) <(III) <(IV)
C (III) <(II) <(I) <(IV)
D (IV) <(I) <(III)< (I)

The correct answer is:
(II) <(I) <(III) <(IV)

Solution:
Resonance and inductive effect decide stability of carbocations.
â´ Correct order of stability is
(II) <(I) <(III) <(IV)

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following cell reaction

2Fes+O2g+4H+aqâ2Fe2+aq+2H2Ol

If Ecell=Ecello at 25oC and Fe2+=10-3M, PO2=0.01 atm and pH=x

Value of x is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
For Ecell=Ecello
log Q must be zero
Q = 1
Q=Fe2+2H+4PO2â10-6(10- 4x)0.01=1
10-4=10-4x
x=1

Topic: Chemistry, Electrochemistry, JEE Main
Consider the following cell reaction
2Fes+O2g+4H+aq→2Fe2+aq+2H2Ol
Eo=1.67 V
At Fe2+=10-3M,PO2=0.1 atm and pH=3 , the cell potential at 25oC is

Options:
A 1.77 V
B 1.57 V
C 1.87 V
D 1.47 V

The correct answer is:
1.57 V

Solution:
2Fes+O2g+4H+aq→2Fe2+aq+2H2O l
N=4 (no. of moles of electrons involved)
From Nernst equation,
Ecell=Ecello-0.0591nlog⁡Q
=1.67-0.05914log log⁡10-320.1×10-34H+=10-pH
=1.67-0.106=1.57 V

Topic: Chemistry, Electrochemistry, JEE Main
Consider the following changes
$\begin{array}{l}
M(s) \longrightarrow M(g) \quad \ldots(1) \\
M(g) \longrightarrow M^{2+}(g)+2 e^{-} \quad \ldots(2) \\
M(g) \longrightarrow M^{+}(g)+e^{-} \quad \ldots(3) \\
M^{+}(g) \longrightarrow M^{2+}(g)+e^{-} \ldots(4) \\
M(g) \longrightarrow M^{2+}(g)+2 e^{-} \ldots(5)
\end{array}$
The second ionisation energy of $M$ could be determined from the energy values associated with

Options:
A $1+2+4$
B $1+5-3$
C $2+3-4$
D $5-3$

The correct answer is:
$5-3$

Solution:
The amount of energy required to take out an electron from the monopositive cation is called second ionisation energy
$\begin{array}{l}
\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} ...(v) \\
\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-} ...(iii)
\end{array}$
On subtracting eq(iii) form eq. (v) we get,
$\mathrm{M} \longrightarrow \mathrm{M}^{2+}+\mathrm{e}^{-}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Consider the following complex ions

PÂ =Â [FeF6]3â, QÂ =Â [V(H2O)Â 6]2+, RÂ =Â [Fe(H2O)6]2+

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Options:
A R < Q < P
B R < P< Q
C Q < R < P
D Q < P < R

The correct answer is:
Q < R < P

Solution:
[FeF6]3– : Fe+3 : [Ar] 3d5

F: Weak field Ligand

No. of unpaired electrons =5

μ=(5(5+2))

μ=35BM

VH2O6+2: V+2: 3d3

No. of unpaired electrons =3

μ=3(3+2)

μ=15BM

FeH2O6+2: Fe+2:3d6

H2O : Weak field Ligand

No. of unpaired electrons =4

μ=4(4+2)

μ=24BM

Hence, the answer is option C.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Consider the following compound A . Select the correct statement.

Options:
A It is more acidic than CH3OH
B It is more acidic than H2SO4
C It does not react with CH3MgBr
D It is a tribasic acid

The correct answer is:
It is more acidic than CH3OH

Solution:
It is a strong dibasic acid and stronger than CH3OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Consider the following compounds :
CH4,CH3-CH3,CH3OH,CH3-CH2-CH3,CH3-CH2-CH2-CH2-CH3

How many compounds can be prepared by Wurtz reaction in high yield?

Options:

The correct answer is:
2

Solution:
Only symmetrical alkanes are prepared using wurtz reaction.
So only CH3âCH3 and CH3âCHCH3|âCHCH3|âCH3 can be prepared using this reaction.

Topic: Chemistry, Hydrocarbons, JEE Main
Consider the following compounds

Order of basicity of these compounds in decreasing order is

Options:
A 4 > 1 > 2 > 3
B 1 > 3 > 4 > 2
C 2 > 3 > 4 > 1
D 1 > 3 > 2 > 4

The correct answer is:
1 > 3 > 2 > 4

Solution:
(i) Basicity â1Electronegativity of nitrogen
(ii) -I group decreases basicity
(iii) Delocalisation decreases basicity.

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following conversion:

Amongst the following, how many methods can be used in any one of steps involved in the mentioned conversion?
i. Hoffmann bromamide degradation
ii. Gabriel phthalimide synthesis
iii. Sandmeyer reaction
iv. Clemmensen reduction
v. Mendius reduction

Options:

The correct answer is:
1

Solution:
The product (amine) contains one C-atom less than the amide. This can be achieved by reaction (i).

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the following data

Heat of combustion of H2(g)=–241.8 kJ mol–1

Heat of combustion of C(s)=–393.5 kJ mol–1

Heat of combustion of C2H5OH(l)=–1234.7 kJ mol–1

The heat of formation of C2H5OH(l) is (–) _____ kJ mol–1 (Nearest integer).

Options:

The correct answer is:
278

Solution:
C2H5OHl→2CO2g+3H2Ol++3O2g

Heat of combustion = Heat of formation of products - Heat of formation of reactants

Heat of combustion of hydrogen is equal to heat formation of water, and heat of combustion of carbon is equal to heat of formation of carbon dioxide.

∆HC=2∆HfoCO2+3∆HfoH2O-∆HfoC2H5OHl

–1234.7=[2×(–393.5)+3×(–241.8)]-∆HfoC2H5OHl

∆HfoC2H5OH=-277.7 kJ/mol

≃-278 kJ/mol

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (06 Apr Shift 2)
Consider the following diagram in which an inextensible string connects two blocks passing via ring. The first block moves with a velocity of 2 m s-1and the second one moves with a velocity of v m s-1. If the ring slides on horizontal rod then the value of v is (in m s-1)

Options:
A 32+1
B 32+1
C 23+1
D 23+1

The correct answer is:
32+1

Solution:
The length of the string will remain constant. Hence,

2 cos30°-1cos30°=vcos45°-1cos45°
∴ v-12=32
∴ v=1+32

Topic: Physics, Laws of Motion, JEE Main
Consider the following diagram. In which direction, the motion of the spool of thread (acted upon by three forces shown) kept over a perfectly rough horizontal surface will take place?

Options:
A Right side
B Left side
C The spool of thread remains in the state of rest
D Information is insufficient

The correct answer is:
Left side

Solution:
Torque about CIR i.e. point of contact is in anticlockwise sense.

Topic: Physics, Rotational Motion, NEET
Consider the following disorders:

Colourblindness, Thalassemia, Klinefelter's syndrome, Myotonic dystrophy, Sickle cell anaemia, Phenylketonuria, Turner's syndrome

How many of the above disorders are Mendelian autosomal disorders?

Options:
A Six
B Two
C Five
D Four

The correct answer is:
Four

Solution:
Mendelian disorder are mainly determined by alternation or mutation in the single gene. Myotonic dystrophy, Sickle cell anaemia, Thalassemia, Phenylketonuria are all Mendelian disorder.

Topic: Biology, Principles of Inheritance and Variation, NEET
Consider the following disorders:

Haemophilia, Sickle cell anemia, Cystic fibrosis, Thalassemia, Phenylketonuria, Polydactyly, Kleinfelter's syndrome, Turner's syndrome

How many of the above disorders are Mendelian Autosomal disorders

Options:
A One
B Two
C Five
D Eight

The correct answer is:
Five

Solution:
The mendelian disorder is mainly determined by alteration or mutation in the single gene. Haemophilia, Sickle cell anemia, Cystic fibrosis, Thalassemia, Phenylketonuria, are all mendelian disorder. Polydactyly, Kleinfelter's syndrome, Turner's syndrome are examples of chromosomal disorders.

Topic: Biology, Principles of Inheritance and Variation, NEET
Consider the following $\mathrm{E}^{\circ}$ values
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}
\end{aligned}
$$
Under standard conditions the potential for the reaction
$\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is

Options:
A
$1.68 \mathrm{~V}$

B
$0.63 \mathrm{~V}$

C
$0.91 \mathrm{~V}$

D
$1.40 \mathrm{~V}$

The correct answer is:
$0.91 \mathrm{~V}$

Solution:
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{RHS}}^{\circ}-\mathrm{E}_{\mathrm{LHS}}^{\circ}$
$=(0.77)-(-0.14)$
$=0.91 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main
Consider the following $\mathrm{E}^{\circ}$ values
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}
\end{aligned}
$$
Under standard conditions the potential for the reaction
$\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is

Options:
A
$1.68 \mathrm{~V}$

B
$0.63 \mathrm{~V}$

C
$0.91 \mathrm{~V}$

D
$1.40 \mathrm{~V}$

The correct answer is:
$0.91 \mathrm{~V}$

Solution:
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{RHS}}^{\circ}-\mathrm{E}_{\mathrm{LHS}}^{\circ}$
$=(0.77)-(-0.14)$
$=0.91 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
Consider the following equation:

2SO2g+O2g⇌2SO3g,ΔH=-190 kJ.

The number of factors which will increase the yield of SO3 at equilibrium from the following is _____ .

A. Increasing temperature
B. Increasing pressure
C. Adding more SO2
D. Adding more O2
E. Addition of catalyst

Options:

The correct answer is:
3

Solution:
The given reaction is exothermic reaction, hence, decreasing temperature is the favourable condition. The change in gaseous moles of products and reactants is negative, hence, increasing pressure is the favourable condition. Addition of reactants and removal of products are the favourable conditions. Catalyst only increase the speed of the reaction.

The yield of SO3 at equilibrium will be due to :
B. Increasing pressure
C. Adding more SO2
D. Adding more O2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Consider the following equations :
2Fe2++H2O2→xA+yB

(in basic medium)

2MnO4-+6H++5H2O2→x'C+y'D+z'E

(in acidic medium)

The sum of the stoichiometric coefficients x,y,x',y' and z' for products A,B,C,D and E respectively, is _______

Options:

The correct answer is:
19

Solution:
(i) 2Fe2++H2O2→2Fe3++2OH−

(ii) 2MnO4-+5H2O2+6H+⟶2Mn2++5O2+8H2O

So sum of x+y+x1+y1+z1=2+2+2+5+8=19

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Consider the following equations:
\( 2 \mathrm{Fe}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{x} \mathrm{A}+\mathrm{yB} \)
(in basic medium)
\( 2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{x}^{\prime} \mathrm{C}+\mathrm{y}^{\prime} \mathrm{D}+\mathrm{z}^{\prime} \mathrm{E} \)
(in acidic medium)
The sum of the stoichiometric coefficients \( \mathrm{x}, \mathrm{y}, \mathrm{x}^{\prime}, \mathrm{y}^{\prime} \) and \( \mathrm{z}^{\prime} \) for products \( \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} \) and \( \mathrm{E} \) respectively, is

Options:

The correct answer is:
19

Solution:
(i) 2Fe2++H2O2→2Fe3++2OH−

(ii) 2MnO4-+5H2O2+6H+⟶2Mn2++5O2+8H2O

So sum of x+y+x1+y1+z1=2+2+2+5+8=19

Topic: Chemistry, Hydrogen, JEE Main
Consider the following equilibrium,
$\begin{aligned}
& 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; K_{\mathrm{G}_1}=2.4 \times 10^{20} \\
& \mathrm{No}(g)+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(g) ; K_{C_2}=1.4
\end{aligned}$
Calculate $K_C$ for the reaction,
$\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)$

Options:
A $8.96 \times 10^{-11}$
B $9.48 \times 10^{-\theta}$
C $8.08 \times 10^{-12}$
D $8.96 \times 10^{11}$

The correct answer is:
$8.96 \times 10^{-11}$

Solution:
For, $2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2$
$\mathrm{K}_{\mathrm{C}_1}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2}=2.4 \times 10^{20}$
For,
$\begin{aligned}
& \text { For, } \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g) \\
& K_{C_2}=\frac{[\mathrm{NOBr}]}{[\mathrm{NO}]\left[\mathrm{Br}_2\right]^{1 / 2}}=1.4 \\
& \text { and } K_C=\frac{[\mathrm{NOBr}]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]^{1 / 2}\left[\mathrm{Br}_2\right]^{1 / 2}} \\
& \text { or } K_C=\sqrt{\frac{1}{K_{\mathrm{C}_1}}} \times K_{C_2}=\sqrt{\frac{1}{2.4 \times 10^{20}}} \times 1.4 \\
& K_C=8.96 \times 10^{-11}
\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the following equilibrium
$$
\mathrm{AgCl} \downarrow+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3 2_2\right]^{+}+\mathrm{Cl}^{-}\right.
$$
White precipitate of $\mathrm{AgCl}$ appears on adding which of the following?

Options:
A
$\mathrm{NH}_3$

B
aqueous $\mathrm{NaCl}$

C
aqueous $\mathrm{HNO}_3$

D
aqueous $\mathrm{NH}_4 \mathrm{Cl}$

The correct answer is:
aqueous $\mathrm{HNO}_3$

Solution:
$2 \mathrm{HNO}_3(\mathrm{aq})+\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}+\mathrm{Cl}^{-}$
$$
\longrightarrow \mathrm{AgCl}(\mathrm{s}) \downarrow+2 \mathrm{NH}_4+2 \mathrm{NO}_3^{-}
$$
When nitric acid is added to amine solution, solution is made acidic and the complex ion dissociates and liberate silver ion to recombine with chloride ion. This is the conformatory test for silver in group 1 .

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2014 (11 Apr Online)
Consider the following first order gas phase reaction at constant temperature
\(\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})\)
If the total pressure of the gases is found to be 200 torr after \(23 \mathrm{sec}\). and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is ______ \(\times 10^{-2} \mathrm{~s}^{-1}\) (nearest integer)
[Given : \(\log _{10}(2)=0.301\)]

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \\ & \mathrm{P}_{23}=\mathrm{P}_0+2 \mathrm{x}=200 \\ & \mathrm{P}_{\infty}=3 \mathrm{P}_0=300 \\ & \mathrm{P}_0=100 \\ & \mathrm{~K}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{\infty}-\mathrm{P}_0}{\mathrm{P}_{\infty}-\mathrm{P}_{\mathrm{t}}} \\ & \mathrm{K}=\frac{2.3}{23} \log \frac{300-100}{300-200} \\ & =\frac{2.3 \times 0.301}{23}=0.0301=3.01 \times 10^{-2} \mathrm{sec}^{-1} \\ & \end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Consider the following for the next items that follow:
The angles $A, B$ and $C$ of a triangle $A B C$ are in the ratio $3: 5: 4$.
What is the value of $a+b+\sqrt{2} c$ equal to?

Options:
A $3 a$
B $2 b$
C $3 b$
D $2 c$

The correct answer is:
$3 b$

Solution:
Given that,
Angles $A, B$ and $C$ of a triangle $A B C$ are in the ratio $3: 5: 4$.
Let the angles $A, B$ and $C$ of a triangle $A B C$ be $3 x, 5 x, 4 x$.
$3 x+5 x+4 x=180$
$12 x=180$
$x=15$
$A, B$ and $C$ of a triangle $A B C$ are $45^{\circ}, 75^{\circ}, 60^{\circ}$
Using sin rule,
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\frac{a}{\sin 45}=\frac{b}{\sin 75}=\frac{c}{\sin 60}$
$\frac{a}{\frac{1}{\sqrt{2}}}=\frac{b}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\frac{c}{\frac{\sqrt{3}}{2}}=k$
$a=\frac{1}{\sqrt{2}} k, b=\frac{\sqrt{3}+1}{2 \sqrt{2}} k, c=\frac{\sqrt{3}}{2} k$
$a+b+\sqrt{2} c=\frac{1}{\sqrt{2}} k+\frac{\sqrt{3}+1}{2 \sqrt{2}} k+\frac{\sqrt{3}}{2} \sqrt{2} k$
$a+b+\sqrt{2} c=\frac{2+\sqrt{3}+1+2 \sqrt{3}}{2 \sqrt{2}} k$
$a+b+\sqrt{ } 2 c=\frac{3+3 \sqrt{3}}{2 \sqrt{2}} k$
$a+b+\sqrt{2} c=\frac{3(1+\sqrt{3})}{2 \sqrt{2}} k$
$\therefore a+b+\sqrt{2} c=3 b$

Topic: Mathematics, Properties of Triangles, JEE Main
Consider the following for the next two $(02)$ items that follow :
In the triangle $A B C$,
$a^2+b^2+c^2=a c+\sqrt{3} b c$
If $\mathrm{c}=8$, what is the area of the triangle?

Options:
A $4 \sqrt{3}$
B $6 \sqrt{3}$
C $8 \sqrt{3}$
D $12 \sqrt{3}$

The correct answer is:
$8 \sqrt{3}$

Solution:
Using cosine rule,
$b^2+c^2-a^2=2 b c \cos A \ldots(i)$
$a^2+c^2-b^2=2 a c \cos B \ldots(ii)$
$a^2+b^2-c^2=2 a b \cos C \ldots(iii)$
Adding (i), (ii) and (iii), we get:
$a^2+b^2+c^2=2 a b \cos C+2 b c \cos A+2 c a \cos B$
According to the question, $a^2+b^2+c^2=a c+\sqrt{3} b c$
Comparing the coefficients of both equations:
$2 \cos C=0 \Rightarrow \cos C=0 \Rightarrow C=90^{\circ}$
$2 \cos B=1 \Rightarrow \cos B=1 / 2 \Rightarrow B=60^{\circ}$
$2 \cos A=\sqrt{3} \Rightarrow \cos A=\sqrt{3} / 2 \Rightarrow A=30^{\circ}$
$\Rightarrow$ The angles of triangles are $\mathrm{A}=30^{\circ}, \mathrm{B}=60^{\circ}, \mathrm{C}=90^{\circ}$
Given, c = 8
$\sin \mathrm{A}=\sin 30^{\circ}=\frac{a}{c}$
$\Rightarrow \frac{1}{2}=\frac{a}{c}$
$\Rightarrow a=4$
$\sin B=\sin 60^{\circ}=\frac{b}{c}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{b}{c}$
$\Rightarrow b=4 \sqrt{3}$
$\therefore$ Area of triangle $=\frac{1}{2} \times \mathrm{a} \times \mathrm{b}$
$=\frac{1}{2} \times 4 \times 4 \sqrt{3}$
$=8 \sqrt{3}$
$\therefore$ Area of the triangle is $8 \sqrt{3}$.

Topic: Mathematics, Properties of Triangles, JEE Main
Consider the following four elements, which are represented according to long form of periodic table

Here W,Y and Z are left, up and right elements with respect to the element ‘X’ and ‘X’ belongs to 16th group and 3rd period. Then according to given information the incorrect statement regarding given elements is

Options:
A Maximum electronegativity : Y
B Maximum catenation property : X
C Maximum electron affinity : Z
D Y exhibits variable covalency

The correct answer is:
Y exhibits variable covalency

Solution:
W : Phosphorus
Electronegativity : O>Cl>S>P
Electron Affinity : Cl>S>O>P
Y : Oxygen

X : Sulphur
Catenation : S>P>O>Cl
Z : Chlorine

Oxygen exhibits covalency of two only

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Consider the following four measures (a - d) that could be taken to successfully grow chick-pea in an area where bacterial blight disease is common
(a) Spray with Bordeaux mixture
(b) Control of the insect vector of the disease pathogen
(c) Use of only disease-free seeds
(d) Use of varieties resistant to the disease
Which two of the above measures can control the disease?

Options:
A (a) and (d)
B (b) and (c)
C (a) and (b)
D (c) and (d)

The correct answer is:
(c) and (d)

Solution:
Bordeaux is used to control fungal diseases and no insect vector is reported for bacterial blight disease. Thus use of disease free seeds and disease resistant varieties can control disease.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2008 (Mains)